CBSE Class 12th Exam Date:17 Feb' 26 - 17 Feb' 26
Have you ever wondered how aeroplanes change direction in the skies or how the forces act in physics? Welcome to the world of vectors, where quantities have both magnitude and direction. Vector algebra deals with mathematical operations on vectors which are having both magnitude and direction, like the addition and subtraction of two vectors, multiplication of two vectors. Also, in this chapter, you will find problems based on angles between two vectors, perpendicular distance between two vectors etc. This article covers all the important class 12 maths chapter 10 question answers based on the latest NCERT syllabus for class 12 and based on Vector Algebra from NCERT Books for class 12 Math.
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These solutions are created by our subject matter experts covering the complete solution of the 10th Chapter of the NCERT Exemplar for Class 12 Mathematics. You will find a healthy number of questions to practice here. Vectors are discussed in detail in NCERT Exemplar Class 12 Maths Solutions Chapter 10.
Class 12 Maths Chapter 10 Exemplar Solutions Exercise: 10.1 Page number: 215-219 Total questions: 45 |
Question:1
Answer:
We have,
$\vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k}$
$\overrightarrow{\mathrm{b}}=2 \hat{\jmath}+\hat{\mathrm{k}}$
Since, unit vector is needed to be found in the direction of the sum of vectors $\vec{a}$ and $\vec{b}$.
So, add vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$.
Let,
$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$
Substituting the values of vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$.
$\Rightarrow \vec{c}=(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})+(2 \hat{\jmath}+\hat{\mathrm{k}})$
$\Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}}+2 \hat{\jmath}+\hat{\mathrm{k}}$
$\Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}$
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)
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,
$\overrightarrow{\mathrm{b}}=2 \hat{\imath}+\hat{\jmath}-2 \hat{\mathrm{k}}$
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)
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 $2 \vec{a}-\vec{b}$, so, we just need to substitute in the above equation.
$
\begin{aligned}
& \Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{|\hat{\jmath}+6 \hat{k}|} \\
& \text { Here },|\hat{\jmath}+6 \hat{k}|=\sqrt{1^2+6^2} \\
& \Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{1^2+6^2}}
\end{aligned}
$
$\begin{aligned} &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{\mathrm{k}}}{\sqrt{1+36}}\\ &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}}\\ &\text { Thus, unit vector in the direction of }\\ &2 \vec{a}-\vec{b} {i s} \frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}} \end{aligned}$
Question:3
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
Answer:
We have been given that,
Position vector of A $=\vec{a}$
$\Rightarrow \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}$
Position vector of $B=\overrightarrow{\mathrm{b}}$
$\Rightarrow \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}$
Here, O is the origin.
We need to find the position vector of C ,
that is, $\overrightarrow{\mathrm{OC}}$
Also, we have
$\overrightarrow{\mathrm{BC}}=1.5 \overrightarrow{\mathrm{BA}} {\ldots(\mathrm{i})}$
Here, $\overrightarrow{\mathrm{BC}}=$ Positionvectorof $\mathrm{C}-$ Position vector of B
$\Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}} \ldots$
And, $\overrightarrow{\mathrm{BA}}=$ Positionvectorof $A-$ Position vector of B
$\Rightarrow \overrightarrow{\mathrm{BA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}$
$\\ \overrightarrow{\mathrm{BC}}=1.5 \overrightarrow{\mathrm{BA}}$
$\Rightarrow \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=1.5(\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}})$
$\Rightarrow \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=1.5 \overrightarrow{\mathrm{OA}}-1.5 \overrightarrow{\mathrm{OB}}$
$\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{OA}}-1.5 \overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OB}}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{OA}}-0.5 \overrightarrow{\mathrm{OB}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{a}}-0.5 \overrightarrow{\mathrm{b}}\\ &[\because \text { it is given that } \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}]\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=\frac{15 \overrightarrow{\mathrm{a}}}{10}-\frac{5 \overrightarrow{\mathrm{b}}}{10} \end{aligned}$
$\Rightarrow \overrightarrow{\mathrm{OC}}=\frac{3 \overrightarrow{\mathrm{a}}}{2}-\frac{\overrightarrow{\mathrm{b}}}{2}$
$\Rightarrow \overrightarrow{\mathrm{OC}}=\frac{3 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}}{2}$
Thus, position vector of point C is $=\frac{3 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}}{2}$
Question:5
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,
$\overrightarrow{\mathrm{OB}}=\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}}$
Position vector of C is given by,
$\vec { OC } = 3 \hat { i } + 5 \hat { j } + 3 \hat { k }$
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 $|\overrightarrow{\mathrm{AB}}|,|\overrightarrow{\mathrm{BC}}|$ 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
$\begin{aligned}& \Rightarrow \sqrt{(3-\mathrm{k})^2+225}\sqrt{40}=\sqrt{(1-\mathrm{k})^2+81} \\& \Rightarrow \sqrt{\left(9+\mathrm{k}^2-6 \mathrm{k}\right)+225}-\sqrt{40}=\sqrt{\left(1+\mathrm{k}^2-2 \mathrm{k}\right)+81}\end{aligned}$
$\left[\because\right.$ by algebraic identity, $\left.(a-b)^2=a^2+b^2-2 a b\right]$
$\begin{aligned}& \Rightarrow \sqrt{\mathrm{k}^2-6 \mathrm{k}+225+9}-\sqrt{40}=\sqrt{\mathrm{k}^2-2 \mathrm{k}+81+1} \\& \Rightarrow \sqrt{\mathrm{k}^2-6 \mathrm{k}+234}-\sqrt{40}=\sqrt{\mathrm{k}^2-2 \mathrm{k}+82}\end{aligned}$
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
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,
$\\ \hat{\mathrm{r}}=l \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}}\\ \Rightarrow \hat{\mathrm{r}}=\pm \frac{1}{\sqrt{3}} \hat{\imath} \pm \frac{1}{\sqrt{3}} \hat{\jmath} \pm \frac{1}{\sqrt{3}} \hat{\mathrm{k}}$
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
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 vector 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
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 r is given by, vector $=6 \times \frac{\vec{r}}{\mid \vec{r}}$ $\Rightarrow$ vector $=6 \times \frac{-\hat{\imath}+2 \hat{\jmath}+2 \hat{k}}{|-\hat{l}+2 \hat{j}+2 \hat{k}|}$
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
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.
Here, and
.
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, parallelograms on the same base and between the same parallels are equal in area, so we have
$|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{b}} \times \vec{\rightarrow} \overrightarrow{\mathrm{c}}| \rightarrow$
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
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
Answer:
Given are points, A, B, C and D.
Let O be the origin.
We have,
Position vector of A
$\\ \begin{aligned} &=\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OA}}=\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}}\\ &\text { Position vector of } \mathrm{B}=2 \hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \end{aligned}$
$\\ \begin{aligned} &\text { Position vector of } c=2 \hat{\imath}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=2 \hat{\mathrm{\imath}}-3 \hat{\mathrm{k}}\\ &\text { Position vector of } D=3 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OD}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned}$
$\begin{aligned} & \text { Now, let us find out } \overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CD}} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=\text { position vector of } B-\text { Position vector of } \mathrm{A} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\hat{\imath}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}-\hat{1}-\hat{\mathrm{j}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\hat{\mathrm{k}} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{\imath}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \text { And, } \overrightarrow{\mathrm{CD}}=\text { position vector of } D-\text { Position vector of } \mathrm{C} \\ & \Rightarrow \overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OC}} \\ & \Rightarrow \overrightarrow{\mathrm{CD}}=(3 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}})-(2 \hat{\imath}-3 \hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{CD}}=3 \hat{\imath}-2 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}}+3 \hat{\mathrm{k}}\end{aligned}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{CD}}=\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}\\ &\text { The projection of } \overrightarrow{\mathrm{AB}} \text { along } \overrightarrow{\mathrm{CD}} \text { is given by, }\\ &\text { Projection }=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}}{|\overrightarrow{\mathrm{CD}}|} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \text { Projection }=\frac{(\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})(\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})}{|\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|}\\ &\Rightarrow \text { Projection }=\frac{(\hat{\imath})(\hat{i})+(-2 \hat{\jmath})(-2 \hat{\jmath})+(4 \hat{\mathrm{k}})(4 \hat{\mathrm{k}})}{\sqrt{1^{2}+(-2)^{2}+4^{2}}}\\ &\text { "we know that, } \hat{i} \times \hat{i}=\hat{\jmath} \times \hat{\jmath}=\hat{\mathrm{k}} \times \hat{\mathrm{k}}=1 \text { and } \hat{i} \times \hat{\mathrm{j}}=\hat{i} \times \hat{\mathrm{k}}=\hat{\mathrm{j}} \times \hat{\mathrm{k}}=0 \end{aligned}$
$\\ \begin{aligned} &\text { So }\\ &\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right)\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\\ &\text { Also, we know that, } \left.|a \hat{\imath}+b \hat{\jmath}+c \hat{k}|=\sqrt{a^{2}+b^{2}+c^{2}}\right]\\ &\Rightarrow \text { Projection }=\frac{1+4+16}{\sqrt{1+4+16}}\\ &\Rightarrow \text { Projection }=\frac{21}{\sqrt{21}} \end{aligned}$
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
Answer:
We have,
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,
$\\ \begin{aligned} &\text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|_{\ldots \text { (i) }}\\ &\text { Let us find out } \overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}} \text { first. } \end{aligned}$
$\\ \begin{aligned} &\text { We can say, }\\ &\text { Position vector of } \mathrm{A}=\hat{i}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OA}}=\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\text { Position vector of } \mathrm{B}=2 \hat{\mathrm{l}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \end{aligned}$
$\\ \begin{aligned} &\text { Position vector of } \mathrm{c}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\overrightarrow{\mathrm{AB}}\\ &\overrightarrow{\mathrm{AB}}=\text { position vector of B-Position vector of } \mathrm{A} \end{aligned}$
$\\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\imath}-\hat{\jmath}+4 \hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AB}}=2 \hat{\imath}-\hat{\imath}-\hat{\jmath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}-3 \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{i}-3 \hat{\jmath}+\hat{\mathrm{k}} \\ \text { For } \overrightarrow{\mathrm{AC}}$
$\\ \begin{aligned} &\overrightarrow{\mathrm{AC}}=\text { position vector of } \mathrm{C} \text { -Position vector of } \mathrm{A}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=(4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=4 \hat{\imath}-\hat{\imath}+5 \hat{\jmath}-2 \hat{\jmath}-\hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=3 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\hat{\imath}(12-3)-\hat{\jmath}(-4-3)+\hat{\mathrm{k}}(3+9)\\ &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=9 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\\ &\text { And, }\\ &|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|9 \hat{\imath}+7 \hat{\jmath}+12 \hat{\mathrm{k}}|\\ &\Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{9^{2}+7^{2}+12^{2}} \end{aligned}$
$\\ \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{81+49+144} \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{274}$
$\begin{gathered}\text {From equation }(i) \text {, weget } \\ \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \\ \Rightarrow \text { Areaof } \triangle \mathrm{ABC}=\frac{1}{2} \times \sqrt{274}\end{gathered}$
Thus, area of triangle ABC is $\frac{\sqrt{274}}{2}$ sq units
Question:14
Answer:
We have,
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
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
$\\ \sin \mathrm{A}=\frac{\text { perpendicular }}{\text { hypotenuse }} \text { in } \Delta \mathrm{BAD} \\ \qquad \because \sin \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{c}} \\ \Rightarrow \mathrm{BD}=\mathrm{c} \sin \mathrm{A}$
$\\ \text { And, } A D=c \cos A \\ \qquad \cos A=\frac{\text { base }}{\text { hypotenuse }} \text { in } \Delta B A D \\ \because \\ \Rightarrow \cos A=\frac{A D}{C} \\ \Rightarrow A D=c \cos A$
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
$
\begin{aligned}
& (B C)^2=(B D)^2+(C D)^2 \\
& \Rightarrow a^2=(c \sin A)^2+(C D)^2[\because \text { from the diagram, } B D=\text { csin } A] \\
& \Rightarrow a^2=c^2 \sin ^2 A+(b-\cos A)^2 \\
& \because \text { fromthediagram, } A C=C D+A D \\
& \Rightarrow C D=A C-A D \\
& \Rightarrow C D=b-\cos A] \\
& \Rightarrow a^2=c^2 \sin ^2 A+\left(b^2+(-\cos A)^2-2 b \cos A\right)[\because \text { from algebraic identity, }(a- \\
& \left.b)^2=a^2+b^2-2 a b\right] \\
& \Rightarrow a^2=c^2 \sin ^2 A+b^2+c^2 \cos ^2 A-2 b c \cos A \\
& \Rightarrow a^2=c^2 \sin ^2 A+c^2 \cos ^2 A+b^2-2 b c \cos A \\
& \Rightarrow a^2=c^2\left(\sin ^2 A+\cos ^2 A\right)+b^2-2 b c \cos A \\
& \Rightarrow a^2=c^2+b^2-2 b c \cos ^2\left[\because \text { from trigonometric identity, } \sin ^2 \theta+\cos ^2 \theta=1\right] \\
& \Rightarrow 2 b c \cos A=c^2+b^2-a^2 \\
& \Rightarrow 2 b c \cos A=b^2+c^2-a^2 \\
& \Rightarrow \cos A=\frac{b^2+c^2-a^2}{2 b c}
\end{aligned}
$
Hence proved
Question:16
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,
$\begin{aligned} & \text { Areaof } \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \\ & \text { Here } \overrightarrow{\mathrm{AB}}=\text { Positionvectorof } \mathrm{B}-\text { Positionvectorof } \mathrm{A} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\end{aligned}$
$\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}} \\ \therefore \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \times(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}})| \\ \Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}| \\ {[\because-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}},-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=0]}$
$\Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|_{\ldots(\mathrm{j})}$
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 $\triangle \mathrm{ABC}$ should be equal to 0.
⇒ Area of $\triangle \mathrm{ABC} = 0$
$\\ \Rightarrow \frac{1}{2}|\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}|=0 \\ \Rightarrow \vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}=0$
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, $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ from equation (i) And, $|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|{\text {from equation (i) }}$
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{\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}}{|\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}|}
$
Question:17
Answer:
We have,
Let ABCD be a parallelogram.
In ABCD,
$\mathrm{AB}=\overrightarrow{\mathrm{p}}$
$\mathrm{AD}=\overrightarrow{\mathrm{q}}$ And since, $\mathrm{AD} \| \mathrm{BC}$
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 $\overrightarrow{\mathrm{b}}$ are diagonals of the parallelogramABCD 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
Answer:
Given that,$\begin{aligned} \vec{a} & =\hat{i}+\hat{j}+\hat{k} and \vec{b} & =\hat{j}-\hat{k}\end{aligned}$
$\\ \begin{aligned} &\text { We need to find vector } \overrightarrow{\mathrm{C}} \text { . }\\ &\text { Let } \overrightarrow{\mathrm{c}}=\mathrm{x} \hat{1}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}, \text { where } \mathrm{x}, \mathrm{y}, \mathrm{z} \text { be any scalars. }\\ &\text { Now, for } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{x} \hat{\mathrm{l}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}) \end{aligned}$
$\\ \Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{xi}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ \mathrm{x} & \mathrm{y} & \mathrm{z} \end{array}\right|=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{1}((1)(\mathrm{z})-(1)(\mathrm{y}))-\hat{\jmath}((1)(\mathrm{z})-(1)(\mathrm{x}))+\hat{\mathrm{k}}(1)(\mathrm{y})-(1)(\mathrm{x}))=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-\mathrm{y})-\hat{\mathrm{j}}(\mathrm{z}-\mathrm{x})+\hat{\mathrm{k}}(\mathrm{y}-\mathrm{x})=\hat{\mathrm{j}}-\hat{\mathrm{k}}$
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)
Also, for $\vec{a} \vec{c}=3$
$\begin{aligned}
& \vec{a} \vec{c}=(\hat{i}+\hat{j}+\hat{k}) \cdot(x \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}}) \\
& \Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=3
\end{aligned}$
$\text { since }2(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\mathrm{x}+\mathrm{y}+\mathrm{z}, \text { as } \hat{1} \cdot \hat{\imath}=\hat{\mathrm{j}} \cdot \hat{\mathrm{j}}=\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}=1 \text { and other dot }$
multiplication is zero. We get,
$\begin{array}{lllllll}
x+y+z=3 \ldots(i v) \text { Now, } & \text { add } & \text { equations } & \text { (ii) } \quad \text { and } & \text { (iii), we } & \text { wet } \\
(x-z)+(x-y)=1+1 & & & & & &
\end{array}$
$\begin{aligned}
& \Rightarrow x+x-y-z=2 \quad \text { Add equations } \\
& \Rightarrow 2 x-y-z=2 \ldots(v) \\
& (x+y+z)+(2 x-y-z)=3+2 \\
& \Rightarrow x+2 x+y-y+z-z=5 \\
& \Rightarrow 3 x=5 \\
& \Rightarrow x=\frac{5}{3}
\end{aligned}$
Put value of x in equation (iii), we get Equation (iii)
$\begin{aligned}
& \Rightarrow x-y=1 \Rightarrow \frac{5}{3}-\mathrm{y}=1 \Rightarrow \mathrm{y}=\frac{5}{3}-1 \\
& \Rightarrow \mathrm{y}=\frac{5-3}{3} \\
& \Rightarrow \mathrm{y}=\frac{2}{3}
\end{aligned}$
(iv) and
(v), we get
Put this value of $y$ in equation (i), we get
Equation(i) $\Rightarrow \mathrm{z}-\mathrm{y}=0 \Rightarrow \mathrm{z}-\frac{2}{3}=0 \Rightarrow \mathrm{z}=\frac{2}{3}$
since, $\vec{c}=x \hat{1}+y \hat{j}+z \hat{k}$
By putting the values of x, y and z, we get
$\overrightarrow{\mathrm{c}}=\frac{5}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}+\frac{2}{3} \hat{\mathrm{k}}$
Thus, we have found the vector $\vec{c}$.
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
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
$A(2 \vec{a}-3 \vec{b}){\text { and }} B(\vec{a}+\vec{b})$.
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
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)
Let $\vec{a}=\mathrm{a}_1 \hat{i}+\mathrm{a}_2 \hat{j}+\mathrm{a}_3 \hat{k}$ then
$\begin{aligned}
& \overrightarrow{\mathrm{a}} \times \hat{\mathrm{i}}=\left(\mathrm{a}_1 \hat{\imath}+\mathrm{a}_2 \hat{\jmath}+\mathrm{a}_3 \hat{\mathrm{k}}\right) \times \hat{\mathrm{i}}=\mathrm{a}_1 \hat{\imath} \times \hat{\mathrm{i}}+\mathrm{a}_2 \hat{\jmath} \times \hat{\mathrm{i}}+\mathrm{a}_3 \hat{\mathrm{k}} \times \hat{\mathrm{i}} \\
& \Rightarrow \vec{a} \times \hat{\mathrm{i}}=0-\mathrm{a}_2 \hat{\mathrm{k}}+\mathrm{a}_3 \hat{\mathrm{j}}(\because \hat{\mathrm{i}} \times \hat{\mathrm{i}}=0, \hat{\mathrm{j}} \times \hat{\mathrm{i}}=-\hat{\mathrm{k}}, \hat{\mathrm{k}} \times \hat{\mathrm{i}}=\hat{\mathrm{j}}) \\
& \Rightarrow \overrightarrow{\mathrm{a}} \times \hat{i}=-\mathrm{a}_2 \hat{k}+\mathrm{a}_3 \hat{j}
\end{aligned}$
$\\ \begin{aligned} &\Rightarrow|\vec{a} \times \hat{i}|^{2}=a_{2}^{2}+a_{3}^{2}\\ &\text { Similarly, we get }\\ &\Rightarrow|\vec{a} \times \hat{\jmath}|^{2}=a_{1}^{2}+a_{3}^{2}\\ &\Rightarrow|\overrightarrow{\mathrm{a}} \times \hat{\mathrm{k}}|^{2}=\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}\\ &\therefore|\vec{a} \times \hat{\imath}|^{2}+|\vec{a} \times \hat{\jmath}|^{2}+|\vec{a} \times \hat{k}|^{2}=a_{2}^{2}+a_{3}^{2}+a_{1}^{2}+a_{3}^{2}+a_{1}^{2}+a_{2}^{2} \end{aligned}$
$=2\left(\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\mathrm{a}_{3}^{2}\right)=2|\overrightarrow{\mathrm{a}}|^{2}\left(\because|\overrightarrow{\mathrm{a}}|=\sqrt{\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\mathrm{a}_{3}^{2}}\right)$
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
$\\A. \lambda = -2\\ B. \lambda = 0\\ C. \lambda = 1\\ D. \lambda = -1\\$
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 $\begin{aligned} &\left|\begin{array}{ccc} \lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda \end{array}\right|=0\\ \end{aligned}$
$\begin{aligned} & \Rightarrow \lambda\left(\lambda^2-1\right)-1(\lambda+2)+2(-1-2 \lambda)=0 \\ & \Rightarrow \lambda^3-\lambda-\lambda-2-2-4 \lambda=0 \\ & \Rightarrow \lambda^3-6 \lambda-4=0 \\ & \Rightarrow(\lambda+2)\left(\lambda^2-2 \lambda-2\right)=0\end{aligned}$
$\\ \Rightarrow \lambda=-2 \text { and } \lambda=\frac{2 \pm \sqrt{(-2)^{2}-4 \times 1 \times-2}}{2}=\frac{2 \pm \sqrt{12}}{2} \\ \Rightarrow \lambda=-2 \text { and } \lambda=1 \pm \sqrt{3}$
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
$\begin{aligned} & \text { A. }\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b} \\ & \text { B. } \frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|} \\ & \text { C. } \frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|} \\ & \text { D. }\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \hat{b}\end{aligned}$
Answer:
Answer :(A)
Let θ be the angle between $\vec {a} $ and $\vec{b}$
From figure we can see that, length 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
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
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
2 & 1 & 2 \\
0 & 1 & 1
\end{array}\right|=\hat{\imath}(1-2)-\hat{\jmath}(2-0)+\hat{\mathrm{k}}(2-0)=-\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}$
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
$\begin{aligned}
& \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|} \\
& \Rightarrow \frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|}=\frac{1}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}-\frac{2}{3} k
\end{aligned}$
Hence, there are two unit length perpendicular to the $\vec{a}$ and $\vec{b}$.
Question:34
Answer:
Let $\vec{a}$ and $\vec{b}$ are two non-collinear vectors.
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
Answer:
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 $\overrightarrow{\mathrm{r}}$
$\Rightarrow\vec{r}\text{ is perpendicular to } \vec{a}, \vec{b}$ and $\vec{c}$
$\Rightarrow \vec{a}, \vec{b}$ and $\vec{c}{\text { are coplanar }}$.
$\Rightarrow \vec{a} \cdot(\vec{b} \times \vec{c})=0$
Question:36
Answer:
Given that , $\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}-2 \hat{\mathrm{k}}$
Let $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ are two diagonals of parallelogram.
$\begin{aligned}
& \Rightarrow \overrightarrow{d_1}=\vec{a}+\vec{b}=(3 \hat{\imath}-2 \hat{\jmath}+2 \hat{k})+(-\hat{\imath}-2 \hat{k})=(3-1) \hat{\imath}+(-2+0) \hat{\jmath}+(2-2) \hat{k} \\
& \Rightarrow \overrightarrow{d_1}=2 \hat{\imath}-2 \hat{j} \\
& \Rightarrow \overrightarrow{d_2}=\vec{a}-\vec{b}=(3 \hat{\imath}-2 \hat{\jmath}+2 \hat{k})-(-\hat{\imath}-2 \hat{k})=(3+1) \hat{\imath}+(-2-0) \hat{\jmath}+(2+2) \hat{k} \Rightarrow \\
& \overrightarrow{d_2}=4 \hat{\imath}-2 \hat{\jmath}+4 \hat{k}
\end{aligned}$
Let $\theta$ be the angle between diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$.
Then, $\overrightarrow{d_1} \cdot \overrightarrow{d_2}=\left|\overrightarrow{d_1}\right|\left|\overrightarrow{d_2}\right| \cos \theta$
$\Rightarrow \cos \theta=\frac{\overrightarrow{d_1} \cdot \overrightarrow{d_2}}{\left|\overrightarrow{d_1}\right|\left|\overrightarrow{d_2}\right|}$
$\\ \cos \theta=\frac{(2 \hat{\imath}-2 \hat{\jmath}) \cdot(4 \hat{i}-2 \hat{\jmath}+4 \hat{k})}{\sqrt{2^{2}+2^{2}} \sqrt{4^{2}+(-2)^{2}+4^{2}}}=\frac{8+4}{\sqrt{8} \sqrt{16+4+16}}(\because \hat{\imath} . \hat{\jmath}=0, \hat{\jmath} \cdot \hat{k}=0, \hat{k} \cdot \hat{\imath}=0) \\ \Rightarrow \cos \theta=\frac{12}{2 \sqrt{2} .6}=\frac{1}{\sqrt{2}} \\ \Rightarrow \theta=\frac{\pi}{4}\left(\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right)$
Question:37
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:38
Answer:
$\\ \text { We have, }|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+(\vec{a} \cdot \vec{b})^{2} \\ =|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right)+(\vec{a} \cdot \vec{b})^{2} \\ =|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta+(\vec{a} \cdot \vec{b})^{2}$
$\\ =|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}+(\vec{a} \cdot \vec{b})^{2} \\ =|\vec{a}|^{2}|\vec{b}|^{2} \\ \text { Thus, }|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}$
Question:39
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:40
Answer:
$\\ \begin{aligned} &\text { Let } \vec{a}=a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\\ &\text { Now, taking dot product of } \vec{a} \text { with } \hat{\imath}, \text { we get }\\ &\vec{a} . \hat{\imath}=\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right) . \hat{\imath}=a_{1} \hat{\imath} . \hat{\imath}+a_{2} \hat{\jmath} . \hat{\imath}+a_{3} \hat{k} . \hat{\imath}\\ &\Rightarrow \vec{a} . \hat{\imath}=a_{1} \hat{\imath} . \hat{\imath}+a_{2} \cdot 0+a_{3} \cdot 0(\because \hat{\jmath} \cdot \hat{\imath}=\hat{k} \cdot \hat{\imath}=0) \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \vec{a} . \hat{\imath}=a_{1} \hat{\imath} . \hat{\imath}+a_{2} .0+a_{3} .0(\because \hat{\jmath} . \hat{\imath}=\hat{k} . \hat{\imath}=0)\\ &\Rightarrow \vec{a} \cdot \hat{\imath}=a_{1}\\ &\text { Similarly, taking dot product of } \vec{a} \text { with } \hat{\jmath} \text { and } \hat{k} \text { , we get }\\ &\vec{a} . \hat{\jmath}=a_{2} \text { and } \vec{a} . \hat{k}=a_{3} \\&\Rightarrow(\vec{a} \cdot \hat{\imath}) \hat{\imath}+(\vec{a} . \hat{\jmath}) \hat{\jmath}+(\vec{a} . \hat{k}) \hat{k}=a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}=\vec{a} \end{aligned}$
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
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:44
Answer:
False
Explanation:
$\\ (\vec{a}+\vec{\mathrm{b}})^{2}=(\vec{\mathrm{a}}+\vec{\mathrm{b}}) \cdot(\vec{\mathrm{a}}+\vec{\mathrm{b}}) \\ \Rightarrow \vec{\mathrm{a}}^{2}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{b}}^{2} \\ \Rightarrow \vec{\mathrm{a}}^{2}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}}^{2}(\because \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=\vec{\mathrm{b}} . \vec{\mathrm{a}}) \\ \Rightarrow \vec{\mathrm{a}}^{2}+2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}}^{2}$
Question:45
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.
The sub-topics that are covered under the Class 12 Maths NCERT exemplar solutions chapter 10 Vector Algebra are:
Frequently Asked Questions (FAQs)
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