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NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra

NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:13 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 10 Vector Algebra helps in answering questions like What is your weight? How should a footballer hit the ball to score a goal? The answers to both these questions would be very different. The answer to the former could be 50 kg, a quantity which has the presence of only one value. The second question's answer is quantities that consist of muscle strength, the direction of the foot and the ball, the air current, and the other players' distance. Such quantities are called Vectors and are discussed in detail in NCERT exemplar Class 12 Maths solutions chapter 10.

Question:1

Find the unit vector in the direction of sum of vectors \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \quad \text { and } \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{j}}+\hat{\mathrm{k}}

Answer:

We have,
https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370407192394.png

https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370407953826.png
Since, unit vector is needed to be found in the direction of the sum of vectors https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370408717444.png and https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370409458743.png .
So, add vectors https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370410214789.png and https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370410957395.png .
Let,
https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370411710338.png
Substituting the values of vectors https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370412468682.png and https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370413235486.png .
\\ \Rightarrow \vec{c}=(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})+(2 \hat{\jmath}+\hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}}+2 \hat{\jmath}+\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}-\hat{\jmath}+2 \hat{\jmath}+\hat{\mathrm{k}}+\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}
We know that a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
\\ \begin{aligned} &\hat{c}=\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{c}}|}\\ &\text { Substitute the value of } \overrightarrow{\mathrm{c}} \text { . }\\ &\Rightarrow \hat{c}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{|2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}|}\\ &\text { Here, }|2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}|=\sqrt{2^{2}+1^{2}+2^{2}} \end{aligned}
\\ \Rightarrow \hat{c}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{2^{2}+1^{2}+2^{2}}} \\ \Rightarrow \hat{\mathrm{c}}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{4+1+4}} \\ \Rightarrow \hat{c}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{9}} \\ \Rightarrow \hat{c}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{3}
Thus, unit vector in the direction of sum of vectors \vec{a}{\text { and }} \vec{b} is \frac{2 \hat{i}+\hat{j}+2 \hat{k}}{3} .

Question:2(i)

If \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}} find the unit vector in the direction of 6 \overrightarrow{\mathrm{b}}

Answer:

We have, \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}
(i). We need to find the unit vector in the direction of 6 \overrightarrow{\mathrm{b}} .
First, let us calculate 6 \overrightarrow{\mathrm{b}} .
As we have,
\\ $\overrightarrow{\mathrm{b}}=2 \hat{\imath}+\hat{\jmath}-2 \hat{\mathrm{k}}$\\ \text{Multiply it by 6 on both sides.} $$\\ \Rightarrow 6 \overrightarrow{\mathrm{b}}=6(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) $$ We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.
\Rightarrow 6 \overrightarrow{\mathrm{b}}=12 \hat{\imath}+6 \hat{\jmath}-12 \hat{\mathrm{k}}
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
6 \hat{\mathrm{b}}=\frac{6 \overrightarrow{\mathrm{b}}}{|6 \overrightarrow{\mathrm{b}}|}
Now we know the value of 6 \overrightarrow{\mathrm{b}} , so just substitute the value in the above equation.
\Rightarrow 6 \hat{\mathrm{b}}=\frac{12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}}{|12 \hat{\mathrm{l}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}|} \\ \text { Here, }|12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}|=\sqrt{12^{2}+6^{2}+(-12)^{2}} \\ \Rightarrow 6 \hat{\mathrm{b}}=\frac{12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}}{\sqrt{144+36+144}} \\ \Rightarrow 6 \hat{\mathrm{b}}=\frac{12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}}{\sqrt{324}}
\\ \begin{aligned} &\Rightarrow 6 \hat{\mathrm{b}}=\frac{12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}}{18}\\ &\text { Let us simplify. }\\ &\Rightarrow 6 \hat{\mathrm{b}}=\frac{6(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{18}\\ &\Rightarrow 6 \hat{\mathrm{b}}=\frac{2 \hat{\mathrm{\imath}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{3} \end{aligned}
Thus, unit vector in the direction of 6 \overrightarrow{\mathrm{b}} is \frac{2 \hat{\mathrm{\imath}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{3}

Question:2(ii)

If a = \hat { i } + \hat { j } + 2 \hat { k } \text { and } b = 2 \hat { i } + \hat { j } - 2 \hat { k } \\ find the unit vector in the direction of 2 \vec { a } - \vec { b }

Answer:

We need to find the unit vector in the direction of 2 \vec { a } - \vec { b }
First, let us calculate .2 \vec { a } - \vec { b }
As we have,
\begin{array}{l} \overrightarrow{\mathrm{a}}=\hat{1}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\ldots}(\mathrm{a}) \\ \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{\imath}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}_{\ldots(\mathrm{b})} \end{array}
Then multiply equation (a) by 2 on both sides,
2 \vec { a } = 2 ( \hat { \imath } + \hat { \jmath } + 2 \hat { k } )
We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.
\Rightarrow 2 \vec { a } = 2 \hat { \imath } + 2 \hat { \jmath } + 4 \hat { k } $ \ldots $ (c)\\
Subtract (b) from (c). We get
\Rightarrow 2 \vec { a } - \vec { b } = 2 \hat { l } - 2 \hat { l } + 2 \hat { j } - \hat { j } + 4 \hat { k } + 2 \hat { k } \\
\Rightarrow 2 \vec{a}-\vec{b}=\hat{\jmath}+6 \hat{k}
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
\\ 2 \hat{a}-\hat{b}=\frac{2 \vec{a}-\vec{b}}{|2 \vec{a}-\vec{b}|}$ \\Now we know the value of $2 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}},\\$ so we just need to substitute in the above equation.\\ $\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{|\hat{\jmath}+6 \hat{k}|}$\\ Here, $|\hat{\jmath}+6 \hat{\mathrm{k}}|=\sqrt{1^{2}+6^{2}}$ \\$\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{1^{2}+6^{2}}}$ \begin{aligned} &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{\mathrm{k}}}{\sqrt{1+36}}\\ &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}}\\ &\text { Thus, unit vector in the direction of }\\ &2 \vec{a}-\vec{b}_{i s} \frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}} \end{aligned}

Question:3

Find a unit vector in the direction of \bar{PQ} , where P and Q have co-ordinates (5, 0, 8) and (3, 3, 2), respectively.

Answer:

We have,
Coordinates of P is (5, 0, 8).
Coordinates of Q is (3, 3, 2).
So,
Position vector of P is given by,
\\ \begin{aligned} &\overrightarrow{\mathrm{OP}}=5 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OP}}=5 \hat{\mathrm{i}}+8 \hat{\mathrm{k}}\\ &\text { Position vector of } Q \text { is given by, }\\ &\overrightarrow{\mathrm{OQ}}=3 \hat{\imath}+3 \hat{\jmath}+2 \hat{\mathrm{k}} \end{aligned}
To find unit vector in the direction of PQ, we need to find position vector of PQ.
Position vector of PQ is given by,
\\ \overrightarrow{\mathrm{PQ}}=\text { Position vector of } \mathrm{Q}-\text { Position vector of } \mathrm{P} \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}} \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})-(5 \hat{\mathrm{l}}+8 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}-8 \hat{\mathrm{k}}
\Rightarrow \overrightarrow{\mathrm{PQ}}=-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}
We know that a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:

\\ \widehat{\mathrm{PQ}}=\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}}{|-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}|} \\ \text { Here, }|-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}|=\sqrt{(-2)^{2}+3^{2}+(-6)^{2}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{\sqrt{(-2)^{2}+3^{2}+(-6)^{2}}}
\\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}}{\sqrt{4+9+36}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\mathrm{l}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{\sqrt{49}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\mathrm{l}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}
Thus, unit vector in the direction of PQ is \frac{-2 \hat{\mathrm{l}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}

Question:4

If \vec{a} and \vec{b} are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA.

Answer:

We have been given that,
Position vector of A =\vec{a}
\begin{aligned} &\Rightarrow \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}\\ &\text { Position vector of } B=\overrightarrow{\mathrm{b}} \end{aligned}\\ $$ \Rightarrow \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}} $$\\$Here, O is the origin.\\ \text{We need to find the position vector of } $\mathrm{C},\\$ \text{that is}, $\overrightarrow{\mathrm{OC}}$\\ \text{Also, we have}\\$\overrightarrow{\mathrm{BC}}=1.5 \overrightarrow{\mathrm{BA}}_{\ldots .(\mathrm{i})}$\\ Here, $\overrightarrow{\mathrm{BC}}=$ $Position vector of $\mathrm{C}$ -$ $Position vector of $\mathrm{B}$\\ $\Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}$...(2)\\ And, $\overrightarrow{\mathrm{BA}}=$ $ Position vector of A-Position vector of $\mathrm{B}$\\ $\Rightarrow \overrightarrow{\mathrm{BA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}_{\ldots(\mathrm{iii})}$ \\ \overrightarrow{\mathrm{BC}}=1.5 \overrightarrow{\mathrm{BA}} \\ \Rightarrow \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=1.5(\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}) \\ \Rightarrow \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=1.5 \overrightarrow{\mathrm{OA}}-1.5 \overrightarrow{\mathrm{OB}} \\ \Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{OA}}-1.5 \overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OB}}
\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{OA}}-0.5 \overrightarrow{\mathrm{OB}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{a}}-0.5 \overrightarrow{\mathrm{b}}\\ &[\because \text { it is given that } \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}]\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=\frac{15 \overrightarrow{\mathrm{a}}}{10}-\frac{5 \overrightarrow{\mathrm{b}}}{10} \end{aligned}
\\ \Rightarrow \overrightarrow{\mathrm{OC}}=\frac{3 \overrightarrow{\mathrm{a}}}{2}-\frac{\overrightarrow{\mathrm{b}}}{2} \\ \Rightarrow \overrightarrow{\mathrm{OC}}=\frac{3 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}}{2}
Thus, position vector of point C is =\frac{3 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}}{2}

Question:5

Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.

Answer:

Let the points be A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3).
Let us find the position vectors of these points.
Assume that O is the origin.
Position vector of A is given by,
\\ \overrightarrow{\mathrm{OA}}=\mathrm{ki}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ $Position vector of $B$ is given by, \\ \overrightarrow{\mathrm{OB}}=\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}}\\ \text{Position vector of } $\mathrm{C}$ is given by, \vec { OC } = 3 \hat { l } + 5 \hat { j } + 3 \hat { k }
Know that, two vectors are said to be collinear, if they lie on the same line or parallel lines.
Since, A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3) are collinear, we can say that:
Sum of modulus of any two vectors will be equal to the modulus of third vector.
This means, we need to find | \vec { AB } |
To find : | \vec { AB } |
Position vector of B-Position vector of A
\Rightarrow \vec { AB } = \vec { OB } - \vec { OA }
\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{AB}}=(\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}})-(\mathrm{k} \hat{\imath}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{AB}}=\hat{\imath}-k \hat{\imath}-\hat{\jmath}+10 \hat{\jmath}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{AB}}=(1-\mathrm{k}) \hat{\imath}+9 \hat{\mathrm{j}}\\ &\text { Now, }\\ &|\overrightarrow{\mathrm{AB}}|=\sqrt{(1-\mathrm{k})^{2}+9^{2}} \end{aligned}
\\ \begin{aligned} &\Rightarrow|\overrightarrow{\mathrm{AB}}|=\sqrt{(1-\mathrm{k})^{2}+81}_{\ldots(\mathrm{i})}\\ &\text { To find }|\overrightarrow{\mathrm{BC}}|_{:}\\ &\overrightarrow{\mathrm{BC}}=\text { position vector of C-Position vector of } \mathrm{B}\\ &\Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}} \end{aligned}
\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{BC}}=(3 \hat{\imath}+5 \hat{\jmath}+3 \hat{\mathrm{k}})-(\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{BC}}=3 \hat{\imath}-\hat{\imath}+5 \hat{\jmath}+\hat{\jmath}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{BC}}=2 \hat{\imath}+6 \hat{j}\\ &\text { Now, }\\ &|\overrightarrow{\mathrm{BC}}|=\sqrt{2^{2}+6^{2}} \end{aligned}
\\ \begin{aligned} &\Rightarrow|\overrightarrow{\mathrm{BC}}|=\sqrt{4+36}\\ &\Rightarrow|\overrightarrow{\mathrm{BC}}|=\sqrt{40}\\ &\text { To find }\\ &|\overrightarrow{\mathrm{AC}}| \end{aligned}
\overrightarrow{\mathrm{AC}} = Position vector of C-Position vector of A
\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\mathrm{ki}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AC}}=3 \hat{\mathrm{i}}-\mathrm{k} \hat{\mathrm{I}}+5 \hat{\mathrm{j}}+10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=(3-\mathrm{k}) \hat{\mathrm{i}}+15 \hat{\mathrm{j}}
\\ \begin{aligned} &\text { Now, }\\ &|\overrightarrow{\mathrm{AC}}|=\sqrt{(3-\mathrm{k})^{2}+15^{2}}\\ &\Rightarrow|\overrightarrow{\mathrm{AC}}|=\sqrt{(3-\mathrm{k})^{2}+225}...(iii) \end{aligned}
\\ Take,\\ $$ |\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|=|\overrightarrow{\mathrm{AC}}|\\ $$Substitute values of |\overrightarrow{\mathrm{AB}}|,|\overrightarrow{\mathrm{BC}}| \text { and }|\overrightarrow{\mathrm{AC}}| from (i), (ii) and (iii) respectively. We get,$$ \sqrt{(1-\mathrm{k})^{2}+81}+\sqrt{40}=\sqrt{(3-\mathrm{k})^{2}+225} $$\\\\ Or\\ $\Rightarrow \sqrt{(3-\mathrm{k})^{2}+225}-\sqrt{40}=\sqrt{(1-\mathrm{k})^{2}+81}$\\ $\Rightarrow \sqrt{\left(9+\mathrm{k}^{2}-6 \mathrm{k}\right)+225}-\sqrt{40}=\sqrt{\left(1+\mathrm{k}^{2}-2 \mathrm{k}\right)+81}$\\ $\left[\because\right.$ \text{by algebraic identity,} $\left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$\\\\$\Rightarrow \sqrt{\mathrm{k}^{2}-6 \mathrm{k}+225+9}-\sqrt{40}=\sqrt{\mathrm{k}^{2}-2 \mathrm{k}+81+1}$ $$\\ \Rightarrow \sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}-\sqrt{40}=\sqrt{\mathrm{k}^{2}-2 \mathrm{k}+82} $$\\ $Squaring on both sides, $$ \\\Rightarrow\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}-\sqrt{40}\right]^{2}=\left[\sqrt{\mathrm{k}^{2}-2 \mathrm{k}+82}\right]^{2} $$ \\ \begin{aligned} &\Rightarrow\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right]^{2}+[\sqrt{40}]^{2}-2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}]=\mathrm{k}^{2}-2 \mathrm{k}+82\\ &\left[\because \text { by algebraic identity, }(a-b)^{2}=a^{2}+b^{2}-2 a b\right]\\ &\Rightarrow \mathrm{k}^{2}-6 \mathrm{k}+234+40-2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}]=\mathrm{k}^{2}-2 \mathrm{k}+82\\ &\Rightarrow \mathrm{k}^{2}-6 \mathrm{k}+234+40-\mathrm{k}^{2}+2 \mathrm{k}-82=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}] \end{aligned}
\\ \Rightarrow \mathrm{k}^{2}-\mathrm{k}^{2}-6 \mathrm{k}+2 \mathrm{k}+234+40-82=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}] \\ \Rightarrow-4 \mathrm{k}+192=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}] \\ \Rightarrow-4 \mathrm{k}+192=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right] .2[\sqrt{10}] \\ \Rightarrow 4(-\mathrm{k}+48)=4\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{10}]
\\ \begin{aligned} &\Rightarrow-\mathrm{k}+48=\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234} \cdot \sqrt{10}\\ &\text { Again, squaring on both sides, we get }\\ &[48-\mathrm{k}]^{2}=\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234} \cdot \sqrt{10}\right]^{2}\\ &\Rightarrow(48)^{2}+k^{2}-2(48)(k)=\left(k^{2}-6 k+234\right)(10)\left[\because \text { by algebraic identity },(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \end{aligned}
\\ \begin{aligned} &\Rightarrow 2304+\mathrm{k}^{2}-96 \mathrm{k}=10 \mathrm{k}^{2}-60 \mathrm{k}+2340\\ &\Rightarrow 10 k^{2}-k^{2}-60 k+96 k+2340-2304=0\\ &\Rightarrow 9 \mathrm{k}^{2}+36 \mathrm{k}+36=0\\ &\Rightarrow 9\left(\mathrm{k}^{2}+4 \mathrm{k}+4\right)=0\\ &\Rightarrow \mathrm{k}^{2}+4 \mathrm{k}+4=0 \end{aligned}
\\ \Rightarrow \mathrm{k}^{2}+2 \mathrm{k}+2 \mathrm{k}+4=0 \\ \Rightarrow \mathrm{k}(\mathrm{k}+2)+2(\mathrm{k}+2)=0 \\ \Rightarrow(\mathrm{k}+2)(\mathrm{k}+2)=0 \\ \Rightarrow \mathrm{k}=-2 \text { or } \mathrm{k}=-2
Thus, value of k is -2.

Question:6

A vector \overrightarrow{\mathrm{r}} is inclined at equal angles to the three axes. If the magnitude of \overrightarrow{\mathrm{r}} is 2\sqrt3 units, find \overrightarrow{\mathrm{r}} .

Answer:

Given that,
Magnitude of \overrightarrow{\mathrm{r}} = 2\sqrt3
\Rightarrow|\vec{r}|=2 \sqrt{3}
Also, given that
Vector \overrightarrow{\mathrm{r}} is equally inclined to the three axes.
This means, direction cosines of the unit vector \overrightarrow{\mathrm{r}} will be same. The direction cosines are (l, m, n).
\mathrm{l}=\mathrm{m}=\mathrm{n}
The direction cosines of a vector are simply the cosines of the angles between the vector and the three coordinate axes.
We know the relationship between direction cosines is,
\\ l^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}=1 \\ \Rightarrow l^{2}+l^{2}+l^{2}=1[\because \mathrm{l}=\mathrm{m}=\mathrm{n}] \\ \Rightarrow 3.l^{2}=1 \\ \Rightarrow l=\pm \frac{1}{\sqrt{3}}
Also, we know that \overrightarrow{\mathrm{r}} is represented in terms of direction cosines as,
\\ \hat{\mathrm{r}}=l \hat{\mathrm{u}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}}\\ \Rightarrow \hat{\mathrm{r}}=\pm \frac{1}{\sqrt{3}} \hat{\imath} \pm \frac{1}{\sqrt{3}} \hat{\jmath} \pm \frac{1}{\sqrt{3}} \hat{\mathrm{k}}
\\ \text{We are familiar with the formula,} $$ \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|} $$ \\ \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|} \\\text{To find } \overrightarrow{\mathrm{r}} \\ \Rightarrow \overrightarrow{\mathrm{r}}=\hat{\mathrm{r}}|\overrightarrow{\mathrm{r}}| \\ $Substituting values of$ |\overrightarrow{\mathrm{r}}| and \hat{\mathrm{r}}
\\ \begin{aligned} &\overrightarrow{\mathrm{r}}=\left(\pm \frac{1}{\sqrt{3}} \hat{\imath} \pm \frac{1}{\sqrt{3}} \hat{\jmath} \pm \frac{1}{\sqrt{3}} \hat{\mathrm{k}}\right)(2 \sqrt{3})\\ &\Rightarrow \overrightarrow{\mathrm{r}}=\pm \frac{1}{\sqrt{3}}(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}})(2 \sqrt{3})\\ &\Rightarrow \overrightarrow{\mathrm{r}}=\pm 2(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}})\\ &\text { Thus, the value of } \overrightarrow{\mathrm{r}}_{\text {is }} \pm 2(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \end{aligned}

Question:7

A vector \overrightarrow{\mathrm{r}} has magnitude 14 and direction ratios 2, 3, –6. Find the direction cosines and components of \overrightarrow{\mathrm{r}} , given that \overrightarrow{\mathrm{r}} makes an acute angle with x-axis.

Answer:

Given that,
Magnitude of vector \overrightarrow{\mathrm{r}} = 14
\Rightarrow|\overrightarrow{\mathrm{r}}|=14
Also, direction ratios = 2 : 3 : -6
\\ \begin{array}{l} \overrightarrow{\mathrm{a}}=2 \mathrm{k} \\ \overrightarrow{\mathrm{b}}=3 \mathrm{k} \end{array} \\ \overrightarrow{\mathrm{c}}=-6 \mathrm{k}\\
Also $\overrightarrow{\mathrm{r}}$ can be defined as,\overrightarrow{\mathrm{r}}=a \hat{\imath}+\mathrm{b} \hat{\jmath}+c \hat{\mathrm{k}}
Know that, the direction cosines of a vector are the cosines of the angles between the vector and the three coordinate axes.


∴, the direction cosines l, m and n are
\\ \mathrm{l}=\frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{r}}|} \\ \Rightarrow l=\frac{2 \mathrm{k}}{14}[\because]=2 \mathrm{k} \text { and }|\overrightarrow{\mathrm{r}}|=14_{\mathrm{J}} \\ \Rightarrow l=\frac{\mathrm{k}}{7} \\ \mathrm{~m}=\frac{\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{r}}|}
\\ \Rightarrow \mathrm{m}=\frac{3 \mathrm{k}}{14}[\because \overrightarrow{\mathrm{b}}=3 \mathrm{k} \text { and }|\overrightarrow{\mathrm{r}}|=14 \mathrm{~g} \\ \mathrm{n}=\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{r}}|} \\ \Rightarrow \mathrm{n}=-\frac{6 \mathrm{k}}{14}[\because]{\mathrm{c}}=-6 \mathrm{k} \text { and }|\overrightarrow{\mathrm{r}}|=14]
\\ \begin{aligned} &\Rightarrow \mathrm{n}=-\frac{3 \mathrm{k}}{7}\\ &\text { And we know that, }\\ &l^{2}+m^{2}+n^{2}=1\\ &\Rightarrow\left(\frac{\mathrm{k}}{7}\right)^{2}+\left(\frac{3 \mathrm{k}}{14}\right)^{2}+\left(-\frac{3 \mathrm{k}}{7}\right)^{2}=1 \end{aligned}
\\ \Rightarrow \frac{\mathrm{k}^{2}}{49}+\frac{9 \mathrm{k}^{2}}{196}+\frac{9 \mathrm{k}^{2}}{49}=1 \\ \Rightarrow \frac{4 \mathrm{k}^{2}+9 \mathrm{k}^{2}+36 \mathrm{k}^{2}}{196}=1 \\ \Rightarrow \frac{49 \mathrm{k}^{2}}{196}=1 \\ \Rightarrow 49 \mathrm{k}^{2}=196
\\ \Rightarrow \mathrm{k}^{2}=\frac{196}{49} \\ \Rightarrow \mathrm{k}^{2}=4 \\ \Rightarrow \mathrm{k}=\pm \sqrt{4} \\ \Rightarrow \mathrm{k}=\pm 2
Since, \overrightarrow{\mathrm{r}} makes an acute angle with x-axis, then k will be positive.
\\ \begin{aligned} &\Rightarrow \mathrm{k}=2\\ &\text { The direction cosines are }\\ &l=\frac{\mathrm{k}}{7}=\frac{2}{7} \end{aligned}
\\ \begin{aligned} &\mathrm{m}=\frac{3 \mathrm{k}}{14}=\frac{3 \times 2}{14}=\frac{3}{7}\\ &\mathrm{n}=-\frac{3 \mathrm{k}}{7}=-\frac{3 \times 2}{7}=-\frac{6}{7}\\ &\text { The components of } \overrightarrow{\mathrm{r}} \text { can be found out by, }\\ &\overrightarrow{\mathrm{r}}=\hat{\mathrm{r}} \cdot|\overrightarrow{\mathrm{r}}| \end{aligned}
\\ \Rightarrow \overrightarrow{\mathrm{r}}=(l \hat{\imath}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}})(14) \\ \Rightarrow \overrightarrow{\mathrm{r}}=\left(\frac{2}{7} \hat{\mathrm{r}}+\frac{3}{7} \hat{\mathrm{j}}-\frac{6}{7} \hat{\mathrm{k}}\right)(14) \\ \Rightarrow \overrightarrow{\mathrm{r}}=14 \times \frac{2}{7} \hat{\mathrm{i}}+14 \times \frac{3}{7} \hat{\mathrm{j}}-14 \times \frac{6}{7} \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{r}}=4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}
Thus, the direction cosines (l, m, n) are \left(\frac{2}{7}, \frac{3}{7},-\frac{6}{7}\right) ; and the components of \overrightarrow{\mathrm{r}} are (4,6,-12)

Question:8

Find a vector of magnitude 6, which is perpendicular to both the vectors 2 \hat{i}-\hat{j}+2 \hat{k} and 4\hat{i}-\hat{j}+3 \hat{k}

Answer:

Let the vectors be \\ \vec{a} and \\ \vec{b} , such that
\\ \vec{a}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k} \\ \vec{b}=4 \hat{\imath}-\hat{\jmath}+3 \hat{k}
We need to find a vector perpendicular to both the vectors \\ \vec{a} and \\ \vec{b}

Any vector perpendicular to both \\ \vec{a} and \\ \vec{b} can be given as,
\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{array}\right| \\ \begin{aligned} \Rightarrow \vec{a} \times \vec{b}=\hat{i}((-1)(3)-(2)(-1))-\hat{j}((2)(3)-(2)(4)) \\ +\hat{k}((2)(-1)-(-1)(4)) \end{aligned} \\ \Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=\hat{\imath}(-3+2)-\hat{\jmath}(6-8)+\hat{\mathrm{k}}(-2+4) \\ \Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=-\hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}}
Let
\\ \overrightarrow{\mathrm{r}}=-\hat{\mathrm{l}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}

As we know, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
So,
\\ $A vector of magnitude 6 in the direction of$ \mathrm{r}$ is given by, vector $=6 \times \frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|}$ \\$\Rightarrow$ vector $=6 \times \frac{-\hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}}}{|-\hat{\mathrm{l}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}|}$ \\Here, $|-\hat{\imath}+2 \hat{\jmath}+2 \hat{k}|=\sqrt{(-1)^{2}+(2)^{2}+(2)^{2}}$ \\ \Rightarrow \text { vector }=6 \times \frac{-\hat{1}+2 \hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{(-1)^{2}+(2)^{2}+(2)^{2}}} \\ \Rightarrow \text { vector }=6 \times \frac{(-\hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}})}{\sqrt{1+4+4}} \\ \Rightarrow \text { vector }=6 \times \frac{-\hat{\mathrm{i}}+2 \hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{9}}
\\ \begin{aligned} &\Rightarrow \text { vector }=6 \times \frac{-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3}\\ &\Rightarrow \text { vector }=2 \times(-\hat{1}+2 \hat{\jmath}+2 \hat{\mathrm{k}})\\ &\Rightarrow \text { vector }=-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k}\\ &\text { Thus, required vector is }-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k} \end{aligned}

Question:9

Find the angle between the vectors 2 \hat{i}-\hat{j}+\hat{k} and 3 \hat{i}+4 \hat{j}-\hat{k}.

Answer:

Given:
\vec{a}=2 \hat{i}-\hat{j}+\hat{k} and \vec{b} =3 \hat{i}+4 \hat{j}-\hat{k}
Assume \theta is angle between \vec{a} and \vec{b}.
\\ \cos \theta =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \\ \\ =\frac{(2 \hat{i}-\hat{j}+\hat{k}) \cdot(3 \hat{i}+4 \hat{j}-\hat{k})}{\sqrt{4+1+1} \sqrt{9+16+1}} \\\\ =\frac{6-4-1}{\sqrt{6} \sqrt{26}}=\frac{1}{2 \sqrt{39}} \\\\ \theta =\cos ^{-1} \frac{1}{2 \sqrt{39}}

Question:10

If \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0, \text { show that } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} . Interpret the result geometrically?

Answer:

Given that,
\begin{aligned} &\vec{a}+\vec{b}+\vec{c}=0\\ &\text { Find the value of } \overrightarrow{\mathrm{b}}\\ &\Rightarrow \overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}} \end{aligned}
\\ \text { Take } \vec{a} \times \vec{b} \\ \vec{a} \times \vec{b}=\vec{a} \times(-\vec{a}-\vec{c}) \\ \Rightarrow \vec{a} \times \vec{b}=(-\vec{a} \times \vec{a})-\vec{a} \times \vec{c} \\ \Rightarrow \vec{a} \times \vec{b}=0-\vec{a} \times \vec{c}
\\ {[\because \vec{a} \times \vec{a}=0]} \\ \Rightarrow \vec{a} \times \vec{b}=-\vec{a} \times \vec{c} \\ \Rightarrow \vec{a} \times \vec{b}=\vec{c} \times \vec{a}_{\ldots(i)}
\\ \text { [\% by anti-commutative law, }-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}_{1} \\ \text { Now, take } \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} \text { . } \\ \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=(-\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{c}} \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{c}}
\\ \qquad\left[\because \overrightarrow{\mathrm{C}} \times \overrightarrow{\mathrm{c}}=0\right]. \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-0 \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}...(ii)
\\ \begin{aligned} &\left[\because\right. \text { by anti-commutative law, }-\vec{a} \times \vec{c}=\vec{c} \times \vec{a}_{]}\\ &\text { From equations (i) and (ii), we have }\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}\\ &\Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \end{aligned}
Now, let us interpret the result graphically.
Let there be a parallelogram ABCD.

1
Here, https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370624968879.png and https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370625718817.png .
And AB and AD sides are making angle θ between them.
Area of parallelogram is given by,
Area of parallelogram = Base × Height
So from the diagram, area of parallelogram ABCD can be written as,
\\ \begin{aligned} &\text { Area of parallelogram }=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta\\ &\text { Or, }\\ &\text { Area of parallelogram }=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \end{aligned}
Since, parallelogram on the same base and between the same parallels are equal in area, so we have
\\ $$ |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}| $$\\ $This also implies that, $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}$\\ $ Thus, $ it is represented graphically.

Question:11

Find the sine of the angle between the vectors \overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\text {and }} \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}

Answer:

We have
\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\text {and }} \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}
Let these vectors be represented as,
\begin{array}{l} \overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{b}}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{\mathrm{k}} \end{array}
Comparing the vectors, we get
\begin{array}{l} a_{1}=3, a_{2}=1 \text { and } a_{3}=2 \\ b_{1}=2, b_{2}=-2 \text { and } b_{3}=4 \end{array}
To find sine of the angle between the vectors\vec{a} \text { and } \vec{b}, we can first find out cosine of the angle between them.
Cosine of the angle between \vec{a} \text { and } \vec{b} is given by,
\\\cos \theta=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}| \overrightarrow{\mathrm{b}} \mid} \\ =\frac{(3 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}})(2 \hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})}{|3 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}||2 \hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}|} \\ =\frac{(3 \hat{\imath})(2 \hat{\imath})+(\hat{j})(-2 \hat{\jmath})+(2 \hat{\mathrm{k}})(4 \hat{\mathrm{k}})}{\sqrt{3^{2}+1^{2}+2^{2}} \sqrt{2^{2}+(-2)^{2}+4^{2}}}
\\ \begin{aligned} &\because\text {we know that, } \hat{i} \times \hat{i}=\hat{\jmath} \times \hat{\jmath}=\hat{\mathrm{k}} \times \hat{\mathrm{k}}=1_{\text {and }} \hat{i} \times \hat{\mathrm{j}}=\hat{i} \times \hat{\mathrm{k}}=\hat{\mathrm{j}} \times \hat{\mathrm{k}}=0\\ &\text { So, }\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right)\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\\ &\text { Also, we know that, } \left.|a \hat{\imath}+b \hat{\jmath}+c \hat{k}|=\sqrt{a^{2}+b^{2}+c^{2}}\right] \end{aligned}
\\ =\frac{6-2+8}{\sqrt{9+1+4} \sqrt{4+4+16}} \\ =\frac{12}{\sqrt{14} \sqrt{24}} \\ =\frac{12}{2 \sqrt{14} \sqrt{6}} \\ =\frac{6}{\sqrt{14 \times 6}}
\\ =\frac{6}{\sqrt{84}} \\ =\frac{6}{2 \sqrt{21}} \\ =\frac{3}{\sqrt{21}}
By algebraic identity, we have
\\ \sin \theta=\sqrt{1-\cos ^{2} \theta} \\ =\sqrt{1-\left(\frac{3}{\sqrt{21}}\right)^{2}} \\ =\sqrt{1-\frac{9}{21}} \\ =\sqrt{\frac{21-9}{21}}
\\ =\sqrt{\frac{12}{21}} \\ =\sqrt{\frac{4}{7}} \\ =\frac{2}{\sqrt{7}}
Thus, sine of the angle between the vectors is \frac{2}{\sqrt{7}} .

Question:12

If A, B, C, D are the points with position vectors

\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, 2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}, 2 \hat{\mathrm{i}}-3 \hat{\mathrm{k}}, 3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} respectively, find the projection of \overrightarrow{\mathrm{AB}} along \overrightarrow{\mathrm{CD}} .

Answer:

Given are points, A, B, C and D.
Let O be the origin.
We have,
Position vector of A
\\ \begin{aligned} &=\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OA}}=\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}}\\ &\text { Position vector of } \mathrm{B}=2 \hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \end{aligned}
\\ \begin{aligned} &\text { Position vector of } c=2 \hat{\imath}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=2 \hat{\mathrm{\imath}}-3 \hat{\mathrm{k}}\\ &\text { Position vector of } D=3 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OD}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned}
\\ $Now, let us find out $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$\\ $\overrightarrow{\mathrm{AB}}=$ position vector of B-Position vector of $\mathrm{A}$\\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\hat{\imath}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}-\hat{1}-\hat{\mathrm{j}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{\imath}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\\ $And, $\overrightarrow{\mathrm{CD}}=$ position vector of $D$ -Position vector of $\mathrm{C}$\\ $\Rightarrow \overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OC}}$\\ $\Rightarrow \overrightarrow{\mathrm{CD}}=(3 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}})-(2 \hat{\imath}-3 \hat{\mathrm{k}})$ \\$\Rightarrow \overrightarrow{\mathrm{CD}}=3 \hat{\imath}-2 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}}+3 \hat{\mathrm{k}}$ \\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{CD}}=\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}\\ &\text { The projection of } \overrightarrow{\mathrm{AB}} \text { along } \overrightarrow{\mathrm{CD}} \text { is given by, }\\ &\text { Projection }=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}}{|\overrightarrow{\mathrm{CD}}|} \end{aligned}
\\ \begin{aligned} &\Rightarrow \text { Projection }=\frac{(\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})(\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})}{|\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|}\\ &\Rightarrow \text { Projection }=\frac{(\hat{\imath})(\hat{i})+(-2 \hat{\jmath})(-2 \hat{\jmath})+(4 \hat{\mathrm{k}})(4 \hat{\mathrm{k}})}{\sqrt{1^{2}+(-2)^{2}+4^{2}}}\\ &\text { "we know that, } \hat{i} \times \hat{i}=\hat{\jmath} \times \hat{\jmath}=\hat{\mathrm{k}} \times \hat{\mathrm{k}}=1 \text { and } \hat{i} \times \hat{\mathrm{j}}=\hat{i} \times \hat{\mathrm{k}}=\hat{\mathrm{j}} \times \hat{\mathrm{k}}=0 \end{aligned}
\\ \begin{aligned} &\text { So }\\ &\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right)\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\\ &\text { Also, we know that, } \left.|a \hat{\imath}+b \hat{\jmath}+c \hat{k}|=\sqrt{a^{2}+b^{2}+c^{2}}\right]\\ &\Rightarrow \text { Projection }=\frac{1+4+16}{\sqrt{1+4+16}}\\ &\Rightarrow \text { Projection }=\frac{21}{\sqrt{21}} \end{aligned}
Multiply numerator and denominator by √21.
\\ \begin{aligned} &\Rightarrow \text { Projection }=\frac{21}{\sqrt{21}} \times \frac{\sqrt{21}}{\sqrt{21}}\\ &\Rightarrow \text { Projection }=\frac{21 \times \sqrt{21}}{21}\\ &\Rightarrow \text { Projection }=\sqrt{21}\\ &\text { Thus, projection of } \overrightarrow{\mathrm{AB}} \text { along } \overrightarrow{\mathrm{CD}} \text { is } \sqrt {21} \text { units. } \end{aligned}

Question:13

Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1).

Answer:

We have,
2
The coordinates of points A, B and C are (1, 2, 3), (2, -1, 4) and (4, 5, -1) respectively.
We need to find the area of this triangle ABC.
We have the formula given as,
\\ \begin{aligned} &\text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|_{\ldots \text { (i) }}\\ &\text { Let us find out } \overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}} \text { first. } \end{aligned}
\\ \begin{aligned} &\text { We can say, }\\ &\text { Position vector of } \mathrm{A}=\hat{i}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OA}}=\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\text { Position vector of } \mathrm{B}=2 \hat{\mathrm{l}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \end{aligned}
\\ \begin{aligned} &\text { Position vector of } \mathrm{c}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\overrightarrow{\mathrm{AB}}\\ &\overrightarrow{\mathrm{AB}}=\text { position vector of B-Position vector of } \mathrm{A} \end{aligned}
\\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\imath}-\hat{\jmath}+4 \hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AB}}=2 \hat{\imath}-\hat{\imath}-\hat{\jmath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}-3 \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{i}-3 \hat{\jmath}+\hat{\mathrm{k}} \\ \text { For } \overrightarrow{\mathrm{AC}}
\\ \begin{aligned} &\overrightarrow{\mathrm{AC}}=\text { position vector of } \mathrm{C} \text { -Position vector of } \mathrm{A}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=(4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=4 \hat{\imath}-\hat{\imath}+5 \hat{\jmath}-2 \hat{\jmath}-\hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=3 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}} \end{aligned}
\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\hat{\imath}(12-3)-\hat{\jmath}(-4-3)+\hat{\mathrm{k}}(3+9)\\ &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=9 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\\ &\text { And, }\\ &|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|9 \hat{\imath}+7 \hat{\jmath}+12 \hat{\mathrm{k}}|\\ &\Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{9^{2}+7^{2}+12^{2}} \end{aligned}
\\ \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{81+49+144} \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{274}
$ From equation (i), we get $\\ Area of \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$ \\$\Rightarrow$ Area of $\Delta \mathrm{ABC}=\frac{1}{2} \times \sqrt{274}$Thus, area of yriangle ABC is \frac{\sqrt{274}}{2}{} $ sq units .

Question:14

Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

Answer:

We have,
3
Given:
There are more than 1 parallelogram, and their bases can be taken as common and they are between same parallels.
To Prove:
These parallelograms whose bases are same and are between the same parallel sides have equal area.
Proof:
Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.
Here,
AB || DC and AE || BF
We can represent area of parallelogram ABCD as,
\text { Area of parallelogram } \mathrm{ABCD}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}_{\ldots \text { .. } \mathrm{i} \text { ) }}
Now, area of parallelogram ABFE can be represented as,
Area of parallelogram ABFE
\\ \begin{aligned} &=\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AE}}\\ &=\overrightarrow{\mathrm{AB}} \times(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DE}})\\ &[\because \text { in right-angled } \triangle A D E, \overrightarrow{A E}=\overrightarrow{A D}+\overrightarrow{D E}]\\ &\Rightarrow \text { Area of parallelogram } \mathrm{ABFE}=\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}}+\mathrm{ka})\\ &[\because \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{DE}}=\mathrm{ka}, \text { where } \mathrm{k} \text { is scalar; } \overrightarrow{\mathrm{DE}} \text { is parallel }\\ &\overrightarrow{\mathrm{AB}}_{\text {and }}\\ &\text { hence } \overrightarrow{\mathrm{DE}}=\mathrm{ka}_{1} \end{aligned}
\\ \begin{aligned} &=\vec{a} \times \vec{b}+\vec{a} \times k \vec{a}\\ &=\vec{a} \times \vec{b}+k(\vec{a} \times \vec{a})\\ &[\because \text { a scalar term can be taken out of a vector product] }\\ &=\vec{a} \times \vec{b}+k \times 0\\ &\left[\because \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=0\right]. \end{aligned}
⇒Area of parallelogram ABFE=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} …(ii)
From equation (i) and (ii), we can conclude that
Area of parallelogram ABCD = Area of parallelogram ABFE
Thus, parallelogram on same base and between same parallels are equal in area.
Hence, proved.

Question:15

Prove that in any triangle ABC, \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c} where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.

Answer:

Given:
a, b, c are magnitudes of the sides opposite to the vertices A, B, C respectively.
⇒ AB = c, BC = a and CA = b
To Prove:
In triangle ABC,
\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}
Construction: We have constructed a triangle ABC and named the vertices according to the question.
Note the height of the triangle, BD.
If ∠BAD = A
Then, BD = c sin A
\\ \qquad \sin \mathrm{A}=\frac{\text { perpendicular }}{\text { hypotenuse }} \text { in } \Delta \mathrm{BAD} \\ \qquad \because \sin \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{c}} \\ \Rightarrow \mathrm{BD}=\mathrm{c} \sin \mathrm{A}
\\ \text { And, } A D=c \cos A \\ \qquad \cos A=\frac{\text { base }}{\text { hypotenuse }} \text { in } \Delta B A D \\ \because \\ \Rightarrow \cos A=\frac{A D}{C} \\ \Rightarrow A D=c \cos A
4
Proof:
Here, components of c which are:
c sin A
c cos A
are drawn on the diagram.
Using Pythagoras theorem which says that,
(hypotenuse)2 =(perpendicular)2 + (base)2
Take triangle BDC, which is a right-angled triangle.
Here,
Hypotenuse = BC
Base = CD
Perpendicular = BD
We get,\\ (BC)\textsuperscript{2} = (BD)\textsuperscript{2} + (CD)\textsuperscript{2}\\ \\ $ \Rightarrow $ a\textsuperscript{2} = (c sin A)\textsuperscript{2} + (CD)\textsuperscript{2} [\because $ from the diagram, BD = c sin A]\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + (b - c cos A)\textsuperscript{2}\\ $\because$ from the diagram, AC = CD + AD\\\\ \\$ \Rightarrow $ CD = AC - AD\\ $ \Rightarrow $ CD = b - c cos A]\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + (b\textsuperscript{2} + (-c cos A)\textsuperscript{2} - 2bc cos A) [\because$ from algebraic identity, (a -b)\textsuperscript{2} = a\textsuperscript{2} + b\textsuperscript{2} - 2ab]\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + b\textsuperscript{2} + c\textsuperscript{2} cos\textsuperscript{2} A - 2bc cos A\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + c\textsuperscript{2} cos\textsuperscript{2} A + b\textsuperscript{2} - 2bc cos A\\\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} (sin\textsuperscript{2} A + cos\textsuperscript{2} A) + b\textsuperscript{2} - 2bc cos A\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} + b\textsuperscript{2} - 2bc cos A [\because$ from trigonometric identity, sin\textsuperscript{2} $ \theta $ + cos\textsuperscript{2} $ \theta $ = 1]\\ \\ $ \Rightarrow $ 2bc cos A = c\textsuperscript{2} + b\textsuperscript{2} - a\textsuperscript{2}\\ \\ $ \Rightarrow $ 2bc cos A = b\textsuperscript{2} + c\textsuperscript{2} - a\textsuperscript{2}\\. \\\Rightarrow \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}
Hence proved

Question:16

If \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} determine the vertices of a triangle, show that \frac{1}{2}[\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}] gives the vector area of the triangle. Hence deduce the condition that the three points a, b, c are collinear. Also find the unit vector normal to the plane of the triangle.

Answer:

Let \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} are vertices of a triangle ABC.
Also, we get
Position vector of A=\overrightarrow{\mathrm{a}}
Position vector of B=\overrightarrow{\mathrm{b}}
Position vector of C=\overrightarrow{\mathrm{c}}
We need to show that,
\frac{1}{2}[\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}] gives the vector are of the triangle.
We know that,
Vector area of triangle ABC is given as,
\\ Area of \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$\\ Here, $\overrightarrow{\mathrm{AB}}=$ Position vector of $\mathrm{B}-$ Position vector of $\mathrm{A}$\\ $\Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$\\ $\overrightarrow{\mathrm{AC}}=$ Position vector of $\mathrm{C}-$ Position vector of $\mathrm{A}$\\ \\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}} \\ \therefore \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \times(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}})| \\ \Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}| \\ {[\because-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}},-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=0]}
\Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|_{\ldots(\mathrm{j})}
Thus, shown.
We know that, two vectors are collinear if they lie on the same line or parallel lines.
For \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} to be collinear, area of the ?ABC should be equal to 0.
⇒ Area of ?ABC = 0
\\ \Rightarrow \frac{1}{2}|\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}|=0 \\ \Rightarrow \vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}=0
Thus, this is the required condition for \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} to be collinear.
Now, we need to find the unit vector normal to the plane of the triangle.
Let \vec{\pi} be the unit vector normal to the plane of the triangle.
\\ \overrightarrow{\mathrm{n}}=\frac{\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|}$ \\Note that, $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}_{\text {from equation }(\mathrm{i})}$ \\And, $|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|_{\text {from equation }(\mathrm{i})}$So, \overrightarrow{\mathrm{n}}=\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}$Thus, unit vector normal to the plane of the triangle is \frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}$

Question:17

Show that area of the parallelogram whose diagonals are given by \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}} \text { is } \frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{2} Also find the area of the parallelogram whose diagonals are 2 \hat{i}-\hat{j}+\hat{k} \text { and } \hat{i}+3 \hat{j}-\hat{k}

Answer:

We have,
5
Let ABCD be a parallelogram.
In ABCD,
\\ \mathrm{AB}=\overrightarrow{\mathrm{p}} \\ \mathrm{AD}=\overrightarrow{\mathrm{q}} $$And since, AD || BC$\mathrm{So}, \mathrm{BC}=\overrightarrow{\mathrm{q}}$We need to show that,\\ $ Area of parallelogram $ \mathrm{ABCD}=\frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{2}\\$Where, \overrightarrow{\mathrm{a}} and \overrightarrow{\mathrm{b}} \text{are diagonals of the parallelogram} \mathrm{ABCD}\\ \text{Now, by triangle law of addition, we get} \\\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}


\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{a}}(\mathrm{say})_{\ldots(\mathrm{i})} \\ \Rightarrow \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{AD}} \\ \text { Similarly, } \\ \Rightarrow \overrightarrow{\mathrm{BD}}=-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}
\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{b}}(\mathrm{say}) \ldots(\mathrm{ii})\\ &\text { Adding equations (i) and (ii), we get }\\ &\vec{a}+\vec{b}=(\vec{p}+\vec{q})+(-\vec{p}+\vec{q}) \end{aligned}
\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{q}}\\ &\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{q}}\\ &\Rightarrow \overrightarrow{\mathrm{q}}=\frac{1}{2}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\\ &\text { And, }\\ &\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})-(-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}) \end{aligned}
\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{q}}\\ &\Rightarrow \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{p}}\\ &\Rightarrow \overrightarrow{\mathrm{p}}=\frac{1}{2}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\\ &\text { Now, } \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \text { can be written as, } \end{aligned}
\\ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left(\frac{1}{2}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\right) \times\left(\frac{1}{2}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\right) \\ \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{1}{4}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\ \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{1}{4}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{b}})
\\ \\ \quad\left[\because \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{b}}=0_{\text {and }}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right]
\begin{array}{l} \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{2}{4}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\ \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{1}{2}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \end{array}
We know that,Vector area of parallelogram ABCD is given by,

Area of parallelogram ABCD \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}
\\ =\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \\ =\frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{2}
Hence, shown.
Now, we need to find the area of parallelogram whose diagonals are 2 \hat{\imath}-\hat{\jmath}+\hat{k}_{\text {and }} \hat{\imath}+3 \hat{\jmath}-\hat{k}
We have already derived the relationship between area of parallelogram and diagonals of parallelogram, which is
\\ \begin{aligned} &\text { Area of parallelogram }=\frac{|\vec{a} \times \vec{b}|}{2}\\ &\text { Here, } \vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k}\\ &\text { And, } \overrightarrow{\mathrm{b}}=\hat{\imath}+3 \hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \text { Area of parallelogram }=\frac{|(2 \hat{i}-\hat{j}+\hat{k}) \times(\hat{i}+3 \hat{\jmath}-\hat{k})|}{2} \end{aligned}
\begin{array}{l} =\frac{1}{2}|| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{array}|| \\ =\frac{1}{2}|[\hat{\imath}((-1)(-1)-(1)(3))-\hat{\jmath}((2)(-1)-(1)(1))+\hat{\mathrm{k}}((2)(3)-(-1)(1))]| \\ =\frac{1}{2}|[\hat{\imath}(1-3)-\hat{\jmath}(-2-1)+\hat{\mathrm{k}}(6+1)]| \\ =\frac{1}{2}|-2 \hat{\imath}+3 \hat{\jmath}+7 \hat{\mathrm{k}}| \end{array}
\\ \begin{aligned} &=\frac{1}{2} \sqrt{(-2)^{2}+3^{2}+7^{2}}\\ &=\frac{1}{2} \sqrt{4+9+49}\\ &=\frac{1}{2} \sqrt{62}\\ &\text { Thus, area of required parallelogram is } \frac{1}{2} \sqrt{62} \text { sq units } \end{aligned}

Question:18

If \\ \overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}\\ \overrightarrow{b}=\hat{j}-\hat{k} find a vector\begin{aligned} \\ &\overrightarrow{\mathrm{c}} \text { such that } &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}_{\text {and }} \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3\end{aligned} .

Answer:

Given that,
\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k} \text { and } \overrightarrow{b}=\hat{j}-\hat{k} \\ \begin{aligned} &\text { We need to find vector } \overrightarrow{\mathrm{C}} \text { . }\\ &\text { Let } \overrightarrow{\mathrm{c}}=\mathrm{x} \hat{1}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}, \text { where } \mathrm{x}, \mathrm{y}, \mathrm{z} \text { be any scalars. }\\ &\text { Now, for } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{x} \hat{\mathrm{l}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}) \end{aligned}
\\ \Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{xi}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ \mathrm{x} & \mathrm{y} & \mathrm{z} \end{array}\right|=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{1}((1)(\mathrm{z})-(1)(\mathrm{y}))-\hat{\jmath}((1)(\mathrm{z})-(1)(\mathrm{x}))+\hat{\mathrm{k}}(1)(\mathrm{y})-(1)(\mathrm{x}))=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-\mathrm{y})-\hat{\mathrm{j}}(\mathrm{z}-\mathrm{x})+\hat{\mathrm{k}}(\mathrm{y}-\mathrm{x})=\hat{\mathrm{j}}-\hat{\mathrm{k}}
Comparing Left Hand Side and Right Hand Side, we get
From coefficient of \hat{i} ⇒ z-y = 0 …(i)
From coefficient of \hat{j} ⇒ -(z-x) = 1
⇒ x-z = 1 …(ii)
From coefficient of \hat{k} ⇒ y-x = -1
⇒ x-y = 1 …(iii)
\\ \begin{aligned} &\text { Also, for } \overrightarrow{a} \overrightarrow{c}=3\\ &\overrightarrow{a} \overrightarrow{c}=(\hat{i}+\hat{j}+\hat{k}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})\\ &\Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=3\\ &\text { since }_{2}(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\mathrm{x}+\mathrm{y}+\mathrm{z}, \text { as } \hat{1} . \hat{\imath}=\hat{\mathrm{j}} \cdot \hat{\mathrm{j}}=\hat{\mathrm{k}} . \hat{\mathrm{k}}=1 \text { and other dot }\\ &\text { multiplication is zero. We get, } \end{aligned}\\x + y + z = 3 $ \ldots $ (iv)\\ \\Now, add equations (ii) and (iii), we get\\ (x - z) + (x - y) = 1 + 1\\ \\ $ \Rightarrow $ x + x - y - z = 2\\ \\ $ \Rightarrow $ 2x - y - z = 2 $ \ldots $ (v)\\ \\Add equations (iv) and (v), we get\\ (x + y + z) + (2x - y - z) = 3 + 2\\ \\ $ \Rightarrow $ x + 2x + y - y + z - z = 5\\ \\ $ \Rightarrow $ 3x = 5\\ \\\\ \Rightarrow x=\frac{5}{3}\\ $Put value of x in equation (iii), we get Equation (iii)$\\ \Rightarrow x-y=1\ \Rightarrow \frac{5}{3}-\mathrm{y}=1 $$ \Rightarrow \mathrm{y}=\frac{5}{3}-1 $$\\\begin{aligned} &\Rightarrow \mathrm{y}=\frac{5-3}{3}\\ &\Rightarrow \mathrm{y}=\frac{2}{3}\\ &\text { Put this value of } y \text { in equation (i), we get } \end{aligned}\\ $Equation $ (\mathrm{i}) \Rightarrow \mathrm{z}-\mathrm{y}=0$ $\Rightarrow \mathrm{z}-\frac{2}{3}=0$ $\Rightarrow \mathrm{z}=\frac{2}{3}$\\ since, $\overrightarrow{\mathrm{c}}=\mathrm{x} \hat{1}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\\$ By putting the values of $x, y$ and $z,$ we get\\ \begin{aligned} &\overrightarrow{\mathrm{c}}=\frac{5}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}+\frac{2}{3} \hat{\mathrm{k}}\\ &\text { Thus, we have found the vector } \overrightarrow{\mathrm{c}} \text { . } \end{aligned}

Question:19

The vector in the direction of the vector \hat{i}-2 \hat{j}+2 \hat{k} that has magnitude 9 is
\\ A. \hat{i}-2 \hat{j}+2 \hat{k}\\ B.\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\\ C.3(\hat{i}-2 \hat{j}+2 \hat{k})\\ D. 9(\hat{i}-2 \hat{j}+2 \hat{k})\\

Answer:

C)
Given is the vector \hat{i}-2 \hat{j}+2 \hat{k}
Let this vector be \vec{a} , such that
\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}
Let us first find the unit vector in the direction of this vector \vec{a} .
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
Unit vector in the direction of the vector \vec{a} is given as,
\\ \begin{aligned} &\hat{a}=\frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}\\ &\text { As, we have } \overrightarrow{\mathrm{a}}=\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}\\ &\text { Then, }\\ &|\overrightarrow{\mathrm{a}}|=|\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}|\\ &\Rightarrow|\vec{a}|=\sqrt{1^{2}+(-2)^{2}+2^{2}}\\ &[\because \text { if }|\vec{p}|=|x \hat{\imath}+y \hat{\jmath}+z \hat{k}| \end{aligned}
\\ \Rightarrow|\vec{p}|=\sqrt{x^{2}+y^{2}+z^{2}} \\ \Rightarrow|\vec{a}|=\sqrt{1+4+4} \\ \Rightarrow|\vec{a}|=\sqrt{9} \\ \Rightarrow|\vec{a}|=3
Therefore,
\\ \hat{\mathrm{a}}=\frac{\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3} \\ {\left[\because \overrightarrow{\mathrm{a}}=\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\text {and }}|\overrightarrow{\mathrm{a}}|=3 \right].}
We have found unit vector in the direction of the vector \hat{i}-2 \hat{j}+2 \hat{k} , but we need to find the unit vector in the direction of \hat{i}-2 \hat{j}+2 \hat{k} but also with the magnitude 9.
We have the formula:
Vector in the direction of \vec{a} with a magnitude of 9=9 \times \frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}
\\ \begin{aligned} &=9 \times \widehat{a}\\ &\text { And } \hat{\mathrm{a}}=\frac{\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3} \text { as just found. }\\ &\text { So, }\\ &\Rightarrow \text { Vector in the direction of } \overrightarrow{\mathrm{a}} \text { with a magnitude of } 9=9 \times \frac{\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3} \end{aligned}
\\ \begin{aligned} &=3(\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}})\\ &\text { Thus, vector in the direction of vector }\\ &\hat{\imath}-2 \hat{\jmath}+2 \hat{k}_{\text {and }} \text { has magnitude } 9 \text { is } 3(\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}) \end{aligned}

Question:20

The position vector of the point which divides the join of points2 \vec{a}-3 \vec{b}_{\text {and }} \vec{a}+\vec{b} in the ratio 3 : 1 is
\\A.\frac{3 \vec{a}-2 \vec{b}}{2}$\\ B. $\frac{7 \overrightarrow{\mathrm{a}}-8 \overrightarrow{\mathrm{b}}}{4}$ \\C.$\frac{3 \overrightarrow{\mathrm{a}}}{4}$ \\D. $\frac{5 \vec{a}}{4}$

Answer:

D)
We are given points 2 \vec{a}-3 \vec{b}_{\text {and }} \vec{a}+\vec{b}
Let these points be
A(2 \vec{a}-3 \vec{b})_{a n d} B(\vec{a}+\vec{b}).
Also, given in the question that,
A point divides AB in the ratio of 3 : 1.
Let this point be C.
⇒ C divides AB in the ratio = 3 : 1
We need to find the position vector of C.
We know the position vector of a point C dividing the line segment joining the points P and Q, whose position vectors are p and q in the ratio m : n internally is given by,
\text { Position vector }=\frac{\mathrm{m} \overrightarrow{\mathrm{q}}+\mathrm{n} \overrightarrow{\mathrm{p}}}{\mathrm{m}+\mathrm{n}}
According to the question, here
m : n = 3 : 1
⇒ m = 3 and n = 1
\\ Also, $\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$\\ And $\overrightarrow{\mathrm{p}}=2 \overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}}\\ $Substituting these values in the formula above, we get\\ Position vector of $C=\frac{3(\vec{a}+\vec{b})+1(2 \vec{a}-3 \vec{b})}{3+1}$ $\\=\frac{3 \vec{a}+3 \vec{b}+2 \vec{a}-3 \vec{b}}{4}$. \begin{aligned} &=\frac{3 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}-3 \overrightarrow{\mathrm{b}}}{4}\\ &=\frac{5 \vec{a}}{4}\\ &\text { Thus, position vector of the point is } \frac{5 \vec{a}}{4} \text { . } \end{aligned}

Question:21

The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively is
\\A. -\hat{i}+12 \hat{j}+4 \hat{k}\\ B. 5 \hat{i}+2 \hat{j}-4 \hat{k}\\ C.-5 \hat{i}+2 \hat{j}+4 \hat{k}\\ D.\hat{i}+\hat{j}+\hat{k}

Answer:

C)
Let initial point be A(2,5,0) and terminal point be B(-3,7,4).So, the required vector joining A and B is the vector \vec{AB} .
\\ \Rightarrow \vec{\mathrm{AB}}=(-3-2) \hat{1}+(7-5) \hat{\mathrm{j}}+(4-0) \hat{\mathrm{k}} \\ =-5 \hat{\imath}+2 \hat{\jmath}+4 \hat{\mathrm{k}}

Question:22

The angle between two vectors \vec{a} and \vec{b} with magnitudes \sqrt3 and 4, respectively, and \vec{a}.\vec{b}=2\sqrt3 is

\\A. \frac{\pi}{6}\\\\ B.\frac{\pi}{3}\\\\ C. \frac{\pi}{2}$ $\\\\ D. \frac{5\pi}{2}

Answer:

Answer :(B)
Given that, |\vec{\mathrm{a}}|=\sqrt{3},|\vec{\mathrm{b}}|=4 \text { and } \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=2 \sqrt{3}
Let θ be the angle between vector a and b.
\\ \text { Then, } \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \cos \theta \\ \Rightarrow 2 \sqrt{3}=\sqrt{3} .4 \cos \theta \\ \Rightarrow \quad \cos \theta=\frac{2 \sqrt{3}}{\sqrt{3} .4}=\frac{1}{2} \\ \Rightarrow \theta=\frac{\pi}{3}

Question:23

Find the value of λ such that the vectors \vec{\mathrm{a}}=2 \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+\hat{\mathrm{k}}_{\text {and }} \vec{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} are orthogonal

A. 0
B. 1
C. \frac{3}{2}
D. -\frac{5}{2}

Answer:

D)
Given that, \vec{\mathrm{a}}{\text { and }} \vec{\mathrm{b}} are orthogonal.
\\ \Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=0 \\ \Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=(2 \hat{\imath}+\lambda \hat{\jmath}+\hat{\mathrm{k}}) \cdot(\hat{\imath}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=0 \\ \Rightarrow 2+2 \lambda+3=0(\because \hat{\mathrm{i}} . \hat{\mathrm{j}}=0, \hat{\mathrm{j}} . \hat{\mathrm{k}}=0, \hat{\mathrm{k}} \cdot \hat{\mathrm{i}}=0) \\ \Rightarrow 2 \lambda=-5 \\ \Rightarrow \lambda=-\frac{5}{2}

Question:24

The value of λ for which the vectors 3 \hat{i}-6 \hat{j}+\hat{k} \text{ and } 2 \hat{i}-4 \hat{j}+\lambda \hat{k} are parallel is
A. \frac{2}{3}

B. \frac{3}{2}
C. \frac{5}{2}
D. \frac{2}{5}

Answer:

Answer:(A)
Given that, 3 \hat{i}-6 \hat{j}+\hat{k} \text{ and } 2 \hat{i}-4 \hat{j}+\lambda \hat{k} are parallel
\\ \Rightarrow \frac{3}{2}=\frac{-6}{-4}=\frac{1}{\lambda} \\ \Rightarrow \lambda=\frac{2}{3}

Question:25

The vectors from origin to the points A and B are \vec{\mathrm{a}}=2 \hat{\mathrm{i}}-3\hat{\mathrm{j}}+2\hat{\mathrm{k}} \text { and }\vec{\mathrm{b}}=2\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}} respectively, then the area of triangle OAB is
A. 340
B. \sqrt{25}
C. \sqrt{229}
D. \frac{1}{2}\sqrt{229}

Answer:

Answer :(D)
Given that, vector from origin to the point A, \vec{OA}=2 \hat{\mathrm{i}}-3\hat{\mathrm{j}}+2\hat{\mathrm{k}} and vector from origin to the point B, \vec{OB}=2\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}
\\ \text { Area of } \Delta \mathrm{OAB}=\frac{1}{2}|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}| \\ =\frac{1}{2}|(2 \hat{\mathrm{l}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})|
\\ =\frac{1}{2}\left|\begin{array}{ccc} \hat{1} & \hat{\jmath} & \hat{\mathrm{k}} \\ 2 & -3 & 2 \\ 2 & 3 & 1 \end{array}\right| \\ =\frac{1}{2}|\hat{\imath}(-3-6)-\hat{\jmath}(2-4)+\hat{\mathrm{k}}(6+6)| \\ =\frac{1}{2}|(-9 \hat{\imath}+2 \hat{\jmath}+12 \hat{\mathrm{k}})| \\ =\frac{1}{2} \sqrt{(-9)^{2}+(2)^{2}+(12)^{2}}
\\ =\frac{1}{2} \sqrt{81+4+144} \\ =\frac{1}{2} \sqrt{229}

Question:26

For any vector \vec{a} , the value of (\vec{a} \times \hat{i})^{2}+(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2} is equal to
A. \vec{a} ^2
B. 3\vec{a} ^2
C. 4\vec{a} ^2
D. 2\vec{a} ^2

Answer:

Answer :(D)
\text { Let } \overrightarrow{a}=\mathrm{a}_{1} \hat{i}+\mathrm{a}_{2} \hat{j}+\mathrm{a}_{3} \hat{k} \text { then }\\ \overrightarrow{\mathrm{a}} \times \hat{\mathrm{i}}=\left(\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{\mathrm{k}}\right) \times \hat{\mathrm{i}}=\mathrm{a}_{1} \hat{\imath} \times \hat{\mathrm{i}}+\mathrm{a}_{2} \hat{\jmath} \times \hat{\mathrm{i}}+\mathrm{a}_{3} \hat{\mathrm{k}} \times \hat{\mathrm{i}} \\ \Rightarrow \overrightarrow{a} \times \hat{\mathrm{i}}=0-\mathrm{a}_{2} \hat{\mathrm{k}}+\mathrm{a}_{3} \hat{\mathrm{j}}(\because \hat{\mathrm{i}} \times \hat{\mathrm{i}}=0, \hat{\mathrm{j}} \times \hat{\mathrm{i}}=-\hat{\mathrm{k}}, \hat{\mathrm{k}} \times \hat{\mathrm{i}}=\hat{\mathrm{j}})\\ \Rightarrow \overrightarrow{\mathrm{a}} \times \hat{i}=-\mathrm{a}_{2} \hat{k}+\mathrm{a}_{3} \hat{j} \\ \begin{aligned} &\Rightarrow|\vec{a} \times \hat{i}|^{2}=a_{2}^{2}+a_{3}^{2}\\ &\text { Similarly, we get }\\ &\Rightarrow|\vec{a} \times \hat{\jmath}|^{2}=a_{1}^{2}+a_{3}^{2}\\ &\Rightarrow|\overrightarrow{\mathrm{a}} \times \hat{\mathrm{k}}|^{2}=\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}\\ &\therefore|\vec{a} \times \hat{\imath}|^{2}+|\vec{a} \times \hat{\jmath}|^{2}+|\vec{a} \times \hat{k}|^{2}=a_{2}^{2}+a_{3}^{2}+a_{1}^{2}+a_{3}^{2}+a_{1}^{2}+a_{2}^{2} \end{aligned}
=2\left(\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\mathrm{a}_{3}^{2}\right)=2|\overrightarrow{\mathrm{a}}|^{2}\left(\because|\overrightarrow{\mathrm{a}}|=\sqrt{\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\mathrm{a}_{3}^{2}}\right)

Question:27

\text { If }|\vec{\mathrm{a}}|=10,|\vec{\mathrm{b}}|=2 \text { and } \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=12, \text { then value of }|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|{\text { is }}
A. 5
B. 10
C. 14
D. 16

Answer:

Answer :(D)
Given that, |\vec{\mathrm{a}}|=10,|\vec{\mathrm{b}}|=2 \text { and } \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=12
Let θ be the angle between vector a and b.
\\ \begin{aligned} &\text { Then, }\\ &\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \cos \theta\\ &\Rightarrow 12=10 \times 2 \cos \theta\\ &\Rightarrow \cos \theta=\frac{12}{20}=\frac{3}{5}\\ &\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{1-\frac{9}{25}} \end{aligned}
\\ \Rightarrow \sin \theta=\pm \frac{4}{5} \\ \text { Now, }|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|=|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \sin \theta \\ \Rightarrow|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|=10 \times 2 \times \frac{4}{5}=16

Question:28

The vectors \lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}-\hat{\mathrm{k}}{\text { and }} 2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}} coplanar if

\\A. \lambda = -2\\ B. \lambda = 0\\ C. \lambda = 1\\ D. \lambda = -1\\

Answer:

Answer :(A)
Given that, \lambda \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}, \hat{\imath}+\lambda \hat{\jmath}-\hat{\mathrm{k}} \text { and } 2 \hat{\imath}-\hat{\jmath}+\lambda \hat{\mathrm{k}} \text { are coplanar. }
\\ \begin{aligned} &\text { Let } \vec{a}=\lambda \hat{\imath}+\hat{\jmath}+2 \hat{k}, \vec{b}=\hat{\imath}+\lambda \hat{\jmath}-\hat{k} \text { and } \vec{c}=2 \hat{\imath}-\hat{\jmath}+\lambda \hat{k}\\ &\text { Now, }\\ &\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{c}} \text { are coplanar } \end{aligned}
If \begin{aligned} &\left|\begin{array}{ccc} \lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda \end{array}\right|=0\\ \end{aligned}
\\ $ \Rightarrow $ $ \lambda $ ($ \lambda $ \textsuperscript{2}-1) -1($ \lambda $ +2) +2(-1-2 $ \lambda $ )=0\\ \\ $ \Rightarrow $ $ \lambda $ \textsuperscript{3}- $ \lambda $ -$ \lambda $ -2-2-4 $ \lambda $ =0\\ \\ $ \Rightarrow $ $ \lambda $ \textsuperscript{3}- 6$ \lambda $ -4 =0\\ \\ $ \Rightarrow $ ($ \lambda $ +2)( $ \lambda $ \textsuperscript{2}-2 $ \lambda $ -2)=0\\ \\ \\ \Rightarrow \lambda=-2 \text { and } \lambda=\frac{2 \pm \sqrt{(-2)^{2}-4 \times 1 \times-2}}{2}=\frac{2 \pm \sqrt{12}}{2} \\ \Rightarrow \lambda=-2 \text { and } \lambda=1 \pm \sqrt{3}

Question:29

If \overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}} are unit vectors such that \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0} , then the value of \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{c}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{a}} is
A. 1
B. 3
C. -\frac{3}{2}
D. None of these

Answer:

Answer :(C)
\begin{aligned} &\text { Given that, }\\ &\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{c}} \text { are unit vectors } \Rightarrow \overrightarrow{\mid \mathrm{a}}|=| \overrightarrow{\mathrm{b}}|=\overrightarrow{\mid \mathrm{c}}|=1{\text { and }}\\ &\vec{a}+\vec{b}+\vec{c}=0 \end{aligned}
\\ \begin{aligned} &\Rightarrow(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0\\ &\Rightarrow \vec{\mathrm{a}}^{2}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{a}} \cdot \vec{\mathrm{c}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{b}}^{2}+\vec{\mathrm{b}} . \vec{\mathrm{c}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{c}}^{2}=0\\ &\Rightarrow \vec{\mathrm{a}}^{2}+\vec{\mathrm{b}}^{2}+\vec{\mathrm{c}}^{2}+2(\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{c}}+\vec{c} \cdot \vec{\mathrm{a}})=0\\ \end{aligned}
\\ \begin{aligned} &\Rightarrow 1+1+1+2(\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} . \vec{\mathrm{c}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{a}})=0(\because \vec{\mathrm{a}}, \vec{\mathrm{b}} \text { and } \vec{\mathrm{c}} \text { are unit vectors }) \end{aligned}
\Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{c}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{a}}=-\frac{3}{2}

Question:30

Projection vector of \vec{a} on \vec{b} is

\\A. \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^{2}}\right) \vec{b}$\\ B. $\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}$\\C. $\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|}$\\ D. $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^{2}}\right) \hat{b}$

Answer:

Answer :(A)
6
Let θ be the angle between \vec {a} $ and $\vec{b}
From figure we can see that,\text{length }\mathrm{OL} is the projection of \\\vec{a} on \overrightarrow{\mathrm{b}} and \overrightarrow{\mathrm{O}} \text{is the projection vector of} \overrightarrow{\mathrm{a}} on \overrightarrow{\mathrm{b}} \\ In \Delta OLA, \text{we have}\\ \cos \theta=\frac{O L}{O A}\\ \Rightarrow \mathrm{OL}=\mathrm{OA} \cos \theta
\\ \Rightarrow \mathrm{OL}=|\overrightarrow{\mathrm{a}}| \cos \theta \\ \qquad \mathrm{OL}=|\overrightarrow{\mathrm{a}}|\left\{\frac{(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})}{|\overrightarrow{\mathrm{a}}||\mathrm{b}|}\right\}\left(\because \cos \theta=\frac{(\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}})}{|\overrightarrow{\mathrm{a}}||\mathrm{b}|}\right) \\ \Rightarrow \mathrm{OL}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}
\\ \begin{aligned} &\text { Now, }\\ &\overrightarrow{\mathrm{OL}}=(\mathrm{OL}) \hat{\mathrm{b}}\\ &\Rightarrow \overrightarrow{\mathrm{OL}}=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}\right\} \hat{\mathrm{b}}\\ &\overrightarrow{\mathrm{OL}}=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}\right\} \frac{\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|^{2}}\right\} \overrightarrow{\mathrm{b}} \end{aligned}

Question:31

If \vec{a},\vec{b},\vec{c} are three vectors such that \vec{a}+\vec{b}+\vec{c}=0 and |\overrightarrow{\mathrm{a}}|=2,|\overrightarrow{\mathrm{b}}|=3,|\overrightarrow{\mathrm{c}}|=5 then value of \vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=0 is
A. 0
B. 1
C. –19
D. 38

Answer:

Answer :(C)
Given that, \vec{a}+\vec{b}+\vec{c}=0 and |\overrightarrow{\mathrm{a}}|=2,|\overrightarrow{\mathrm{b}}|=3,|\overrightarrow{\mathrm{c}}|=5
\\ \Rightarrow(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0 \\ \Rightarrow \vec{a}^{2}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{a}+\vec{b}^{2}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+\vec{c}^{2}=0 \\ \Rightarrow \vec{a}^{2}+\vec{b}^{2}+\vec{c}^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0
\\ \Rightarrow 4+9+25+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0(\because \vec{a}|=2,| \vec{b}|=3, \overrightarrow{\mid c}|=5) \\ \Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-\frac{38}{2}=-19

Question:32

If \left | \vec{a} \right |=4 and -3\leq\lambda\leq2, then the range of \left | \lambda a \right | is
A. [0, 8]
B. [–12, 8]
C. [0, 12]
D. [8, 12]

Answer:

Answer :(C)
Given that, \left | \vec{a} \right |=4 and -3\leq\lambda\leq2,
\\ \begin{aligned} &\text { We know that, }|\lambda \vec{\mathrm{a}}|=|\lambda||\vec{\mathrm{a}}|\\ &\Rightarrow|\lambda \vec{\mathrm{a}}|=|-3||\vec{\mathrm{a}}|=3.4=12 \text { at } \lambda=-3\\ &\Rightarrow|\lambda \vec{\mathrm{a}}|=|0||\vec{\mathrm{a}}|=0.4=0 \text { at } \lambda=0\\ &\Rightarrow|\lambda \vec{\mathrm{a}}|=|2||\vec{\mathrm{a}}|=2.4=8 \text { at } \lambda=2\\ &\text { Hence, the range of }|\lambda \vec{\mathrm{a}}| \text { is }[0,12] \end{aligned}

Question:33

The number of vectors of unit length perpendicular to the vectors \vec{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\text { and } \vec{\mathrm{b}}=\hat{\mathrm{j}}+\hat{\mathrm{k}} is
A. one
B. two
C. three
D. infinite

Answer:

Answer :(B)
Given that , \vec{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\text { and } \vec{\mathrm{b}}=\hat{\mathrm{j}}+\hat{\mathrm{k}}
Now, a vector which is perpendicular to both \vec{a} \text{ and } \vec{b} is given by
\\ \vec{a} \times \vec{b}=\left|\begin{array}{lll}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1\end{array}\right|=\hat{\imath}(1-2)-\hat{\jmath}(2-0)+\hat{\mathrm{k}}(2-0)=-\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}$\\ Now, $|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(-2)^{2}+(2)^{2}}=\sqrt{1+4+4}=\sqrt{9}=3$ \\$\therefore$ the required unit vector\\ =\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{-\hat{i}-2 \hat{\jmath}+2 \hat{k}}{3}=\frac{-1}{3} \hat{\imath}-\frac{2}{3} \hat{\jmath}+\frac{2}{3} k$\\ There are two perpendicular directions to any plane. Thus, another unit vector perpendicular to $\vec{a}$ and $\vec{b}$ is given $b y-\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|}$\\ $\Rightarrow \frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|}=\frac{1}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}-\frac{2}{3} k$ Hence, there are two unit length perpendicular to the \vec{a} \text{ and } \vec{b} .

Question:34

Fill in the blanks
The vector \vec{a} + \vec{b} bisects the angle between the non-collinear vectors \vec{a} and \vec{b} if ________

Answer:

Let \vec{a} and \vec{b} are two non-collinear vectors.
7
Let \vec{a} + \vec{b} bisects the angle between \vec{a} and \vec{b} .
\\ \Rightarrow \theta_{1}=\theta_{2} \\ \qquad \cos \theta_{1}=\frac{\vec{a} \cdot(\vec{a}+\vec{b})}{|\vec{a}||\vec{a}+\bar{b}|} \text { and } \cos \theta_{2}=\frac{\vec{b} \cdot(\vec{a}+\vec{b})}{|\vec{b}||\vec{a}+\vec{b}|} \\ \text { since, } \theta_{1}=\theta_{2} \Rightarrow \cos \theta_{1}=\cos \theta_{2} \\ \therefore \quad \frac{\vec{a} \cdot(\vec{a}+\vec{b})}{|\vec{a}||\vec{a}+\vec{b}|}=\frac{\vec{b} \cdot(\vec{a}+\vec{b})}{|\vec{b} \| \vec{a}+\vec{b}|} \\ \Rightarrow |\vec{a}|=|\vec{b}|
Thus, the vector \vec{a} + \vec{b} bisects the angle between the non-collinear vectors \vec{a} and \vec{b} if they are equal.

Question:35

Fill in the blanks
If \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=0, \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{b}}=0, and \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}}=0 for some non-zero vector \overrightarrow{\mathrm{r}} , then the value of \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) is _________

Answer:

8
Given that, \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=0, \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{b}}=0, and \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}}=0 for some non-zero vector \overrightarrow{\mathrm{r}}
\\ \Rightarrow \vec{r} $ is perpendicular to $\vec{a}, \vec{b}$ and $\vec{c}$\\ $\Rightarrow \vec{a}, \vec{b}$ and $\vec{c}$ are coplanar.\\ $\Rightarrow \vec{a} \cdot(\vec{b} \times \vec{c})=0$\\

Question:37

Fill in the blanks
The values of k for which |k \vec{a}|<|\vec{a}| \text { and } k \vec{a}+\frac{1}{2} \vec{a} is parallel to \vec{a} holds true are _______.

Answer:

Given that, |k \vec{a}|<|\vec{a}|
\\ \begin{aligned} &\Rightarrow|k||\vec{a}|<|\vec{a}|\\ &\Rightarrow|k|<1\\ &\Rightarrow-1<k<1\\ &\text { Also, }\\ &k \vec{a}+\frac{1}{2} \vec{a} \text { is parallel to } \vec{a} \end{aligned}
⇒ k cannot be equal to -\frac{1}{2} , otherwise it will become null vector and then it will not be parallel to \vec{a} .
Since, k is along the direction of \vec{a} and not in its opposite direction.
\therefore \mathrm{k} \in(-1,1)-\left\{-\frac{1}{2}\right\}

Question:39

if \left | \vec{a} \times \vec{b} \right |^{2}+\left | (\vec{a} \cdot \vec{b}) \right |^{2}=144 and \left |\vec{a} \right |=4 then \left |\vec{b} \right | is equal to ________.

Answer:

Given that, \left | \vec{a} \times \vec{b} \right |^{2}+\left | (\vec{a} \cdot \vec{b}) \right |^{2}=144 and \left |\vec{a} \right |=4
\\|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144 \\ \Rightarrow|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta=144 \\ \Rightarrow|\vec{a}|^{2}|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=144
\\ \Rightarrow|\vec{a}|^{2}|\vec{b}|^{2}(1)=144 \\ \Rightarrow|\vec{a}|^{2}|\vec{b}|^{2}=144 \\ \Rightarrow|\vec{a}||\vec{b}|=12 \\ \Rightarrow \quad 4 \cdot|\vec{b}|=12 \\ \Rightarrow|\vec{b}|=3

Question:41

True and False
If |\vec{a}|=|\vec{b}| , then necessarily at implies \vec{a}=\pm\vec{b} .

Answer:

False
Explanation:
\\ \begin{aligned} &\text { Let } \vec{a}=\hat{\imath}-2 \hat{\jmath}-3 \hat{k} \text { and } \vec{b}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}\\ &\Rightarrow|\vec{a}|=\sqrt{(1)^{2}+(-2)^{2}+(-3)^{2}}=\sqrt{1+4+9}=\sqrt{14}\\ &\text { and }|\vec{b}|=\sqrt{(1)^{2}+(2)^{2}+(3)^{2}}=\sqrt{1+4+9}=\sqrt{14}\\ &\text { Now, we observe that }\\ &|\vec{a}|=|\vec{b}| \text { but } \vec{a} \neq \vec{b} \end{aligned}

Question:42

True and False
Position vector of a point P is a vector whose initial point is origin.

Answer:

True
Explanation:
Consider a point P in space, having coordinates (x, y, z) with respect to the origin O(0, 0, 0). Then, the vector \vec{OP} having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O.

Question:43

True and False
If |\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| , then the vectors \vec{a} and \vec{b} are orthogonal.

Answer:

True
Explanation:
Given that, |\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|
On squaring both the sides, we get
\\ \Rightarrow|\vec{a}+\vec{b}|^{2}=|\vec{a}-\vec{b}|^{2} \\ \Rightarrow \vec{a}^{2}+2 \vec{a} \cdot \vec{b}+\vec{b}^{2}=\vec{a}^{2}-2 \vec{a} \cdot \vec{b}+\vec{b}^{2} \\ \Rightarrow 2 \vec{a} \cdot \vec{b}=-2 \vec{a} \cdot \vec{b} \\ \Rightarrow 2 \vec{a} \cdot \vec{b}+2 \vec{a} \cdot \vec{b}=0 \\ \Rightarrow 4 \vec{a} \cdot \vec{b}=0\\ \Rightarrow \vec{a} \cdot \vec{b}=0
Hence, \vec{a} and \vec{b} are orthogonal.

Question:45

True and False
If \vec{a}. and \vec{b} are adjacent sides of a rhombus, then \vec{a}.\vec{b}=0 .

Answer:

False
Explanation:
Given that, \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=0 \Rightarrow \vec{\mathrm{a}} \text { and } \vec{\mathrm{b}} are perpendicular to each other.
But, adjacent sides of rhombus are not perpendicular.

More About NCERT Exemplar Class 12 Maths Solutions Chapter 10

NCERT exemplar Class 12 Maths chapter 10 solutions would help achieve academic excellence, be it in the 12 board exams or additional exams, get rid of doubts in fundamental concepts, and promote intellectual curiosity. The students can also use NCERT exemplar Class 12 Maths solutions chapter 10 PDF download, to access the solutions offline.

Sub-topics Covered in NCERT exemplar solutions for Class 12 Maths chapter 10 Vector Algebra

The sub-topics that are covered under the Class 12 Maths NCERT exemplar solutions chapter 10 Vector Algebra are:

  • Basic concepts
  • Directed line
  • Terminal point
  • Magnitude
  • Position Vectors
  • Direction cosines
  • Types of Vectors
  • Zero Vectors
  • Unit Vectors
  • Co-initial Vectors
  • Collinear Vectors
  • Equal Vectors
  • Negative of a Vector
  • Addition of Vectors
  • Properties of Vector Addition
  • Multiplication of a Vector by a Scalar
  • Components of a Vector
  • Vector joining two points
  • Section formula
  • Product of two Vectors
  • Scalar(or dot) product
  • Projection of a Vector on a line
  • Vector(or cross) product

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What can you learn in NCERT Exemplar Class 12 Maths Chapter 10 Solutions?

  • Class 12 Maths NCERT exemplar solutions chapter 10 explore the real-world concepts and applications of Vectors. Vectors have several real-world uses in Aerospace, Fluid concepts, Complex calculations, sports, etc.
  • They have applications in electromagnetism, hydrodynamics, blood flow, calculation of the distance between aircraft in space and the angle between their paths.
  • Their broad range of reach also includes setting up Solar panels considering the roof's location and the direction of the sunbeams.
  • Vectors are used to launch satellites, calculate the distance between their panels, find their prospective trajectory, develop new targeting techniques in warfare, count on-location services like GPS, calculate the ball's trajectory on a field, and a thousand other areas. Knowledge of all the aspects of Vector algebra can help one do well in academics and find out how the world works.
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NCERT Exemplar Class 12 Maths Solutions

Benefits of NCERT Exemplar Class 12 Maths Chapter 10 Solutions

  • NCERT exemplar solutions for Class 12 Maths chapter 10 define Vector algebra as a type of algebra that involves vector form elements (having magnitude and direction), and their algebraic expression follows vector laws.
  • If quantities revolving around velocity, displacement, acceleration, force, weight, spark immediate curiosity in you, Class 12 Maths NCERT exemplar solutions chapter 10 will be fun
  • In this chapter, we will explore some concepts of vectors, the basic algebraic operations of vector and scalar addition and multiplication, their properties related to vector algebra, the geometric properties, the properties to be followed for the proper visualisation of vectors, and their applications and uses in the modern world.
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Frequently Asked Question (FAQs)

1. What are the topics covered in the chapter?

This basic maths chapter covers the topics like vectors, properties, vector by scalar, types of vector etc.

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Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


Best CBSE schools in Delhi: Click Here


In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

  • Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
  • NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

  • Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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