RD Sharma Class 12 Exercise 24.1 Vector or Cross Product Solutions Maths-Download PDF Online

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RD Sharma Class 12 Exercise 24.1 Vector or Cross Product Solutions Maths-Download PDF Online

RD Sharma Class 12 Exercise 24.1 Vector or Cross Product Solutions Maths-Download PDF Online

Satyajeet KumarUpdated on 27 Jan 2022, 06:10 PM IST

RD Sharma’s maths books are well known for their quality and authenticity. Still, many students find it challenging to solve the problems and lag. This problem is that a student is unable to find sources from where they can understand and learn the concepts. RD Sharma Class 12th Exercise 24.1 solutions have arrived to solve this issue with complete reliability on them.

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RD Sharma Class 12 Solutions Chapter 24 Vector or Cross Product - Other Exercise
Vector or Cross Product Excercise: 24.1
RD Sharma Chapter-wise Solutions

The chapter ‘Vector or Cross Product’ solutions are available for access at ease for students. RD Sharma Class 12th Chapter 24 Exercise 24.1 is as per the latest CBSE guidelines to help students score exceptionally in their board examinations. RD Sharma Solutions Cross Products, Unit vectors, area of parallelogram determined by the vectors, area of triangles by vectors and verify laws by vectors are discussed in this chapter. The cross-product of two vectors being the third vector perpendicular to the two original vectors is the concept discussed here. 48 questions are given in this exercise. Class 12th RD Sharma Chapter 24 Exercise 24.1 Solutions are available in PDF format and downloaded for free.

RD Sharma Class 12 Solutions Chapter 24 Vector or Cross Product - Other Exercise

Vector or Cross Product Excercise: 24.1

Vector or Cross Product exercise 24.1 question 1

Answer : $\sqrt{91}$
Hint : To solve this equation by using determinate formula
Given : $\vec{a}=\hat{\imath}+3 \hat{\jmath}-2 \hat{k} \text { and } \vec{b}=\widehat{-i}+3 \hat{k}$
find $\left | \vec{a}\times \vec{b} \right |$
Solution :
$\begin{gathered} \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{c} & \hat{\jmath} & \hat{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{array}\right| \\ =9 \hat{i}-\hat{j}+3 \hat{k} \end{gathered}$
$\begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{9^{2}+(-1)^{2}+(3)^{2}} \\\\ &=\sqrt{81+1+3} \\\\ &=\sqrt{91}\ \end{aligned}$

Vector or Cross Product exercise 24.1 question 2 (i)

Answer : $\sqrt{26}$
Hint : To solve this equation we use determinate formula then magnitude formula
Given : $\vec{a}=3 \hat{\imath}+4 \hat{\jmath}$
$\vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k}$
Find value $|\vec{a} \times \vec{b}|$
Solution :
$\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(4 \times 1-0 \times 1)-\hat{\jmath}(3 \times 1-0 \times 1)+\hat{k}(3 \times 1+4 \times 1) \\\\ &=4 \hat{\imath}-3 \hat{\jmath}-\hat{k} \end{aligned}$
$\begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{4^{2}+3^{2}+1^{2}} \\\\ &=\sqrt{16+9+1} \\\\ &=\sqrt{26} \end{aligned}$

Vector or Cross Product exercise 24.1 question 2 (ii)

Answer :$\sqrt{6}$
Hint : To solve this equation we use determinate formula then magnitude formula
Given : $\vec{a}=2 \hat{\imath}+\hat{k}$
$\begin{aligned} &\vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k} \\ &\text { find }|\vec{a} \times \vec{b}| \end{aligned}$
Solution :
$\begin{aligned} : \vec{a} \times \vec{b} &=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(0 \times 1-1 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2 \times 1-0 \times 1) \\\\ &=-\hat{\imath}-\hat{\jmath}+2 \hat{k} \end{aligned}$
$\begin{aligned} |\vec{a} \times \vec{b}| &=\sqrt{1^{2}+1^{2}+2^{2}} \\\\ &=\sqrt{1+1+4} \\\\ &=\sqrt{6} \end{aligned}$

Vector or Cross Product exercise 24.1 question 3 (i)

Answer :$\frac{1}{3}(-\hat{\imath}+2 \hat{\jmath}+2 \hat{k})$
Hint : To solve this equation , use magnitude and $\vec{a} \times \vec{b}$
Given :$4 \hat{\imath}-\hat{\jmath}+3 \hat{k} ;-2 \hat{\imath}+\hat{\jmath}-2 \hat{k}$
Solution :$\vec{a}=4 \hat{\imath}-\hat{\jmath}+3 \hat{k}$
$\vec{b}=-2 \hat{\imath}+\hat{\jmath}-2 \hat{k}$
$\begin{aligned} \vec{a} \times \vec{b} &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{array}\right| \\\\ &=\hat{i}(1 \times 2-3 \times 1)-\hat{j}(4 \times-2-3 \times-2)+\hat{k}(4 \times 1-(-1 \times-2)) \end{aligned}$
$=-\hat{i}+2 \hat{j}+2 \hat{k}$
$\begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+2^{2}+2^{2}} \\ \end{aligned}$
$\begin{aligned} &=\sqrt{9} \\\\ &=3 \\\\ &\frac{\vec{a} \times \vec{b}}{|\vec{a}+\vec{b}|}=\frac{1}{3}(-\hat{\imath}+2 \hat{\jmath}+2 \hat{k}) \end{aligned}$

Vector or Cross Product exercise 24.1 question 3 (ii)

Answer : $\frac{1}{11}(\hat{\imath}+\hat{\jmath}-3 \hat{k})$
Hint : To solve this equation we use $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$
Given : $\vec{a}=2 \hat{\imath}+\hat{\jmath}+\hat{k}$
$\vec{b}=\hat{\imath}+2 \hat{\jmath}+\hat{k}$
Solution : $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right|$
$\begin{aligned} &=\hat{\imath}(1 \times 1-2 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2 \times 2-1 \times 1) \\\\ &=\hat{\imath}(-1)-\hat{\jmath}(1)+\hat{k}(3) \\\\ &=-\hat{\imath}-\hat{\jmath}+3 \hat{k} \end{aligned}$
$|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(-1)^{2}+3^{2}}$
$\begin{aligned} &=\sqrt{11} \\\\ &\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{1}{\sqrt{11}}(-\hat{\imath}-\hat{\jmath}+3 \hat{k}) \end{aligned}$

Vector or Cross Product exercise 24.1 question 4

Answer : $\sqrt{74}$
Hint : To solve this equation we suppose both terms in x and y then we use magnitude formula
Given :$\vec{a}=(3 \hat{k}+4 \hat{\jmath}) \times(\hat{\imath}+\hat{\jmath}-\hat{k})$
Solution :
$\begin{aligned} &\vec{x}=3 \hat{k}+4 \hat{j} \\\\ &\vec{y}=\hat{\imath}+\hat{\jmath}-\hat{k} \end{aligned}$
$\begin{aligned} &|\vec{x} \times \vec{y}| \\ &\vec{x} \times \vec{y}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 0 & 4 & 3 \\ 1 & 1 & -1 \end{array}\right| \end{aligned}$
$\begin{aligned} &=\hat{\imath}(4 \times-1-3 \times 1)-\hat{\jmath}(0 \times 1-3 \times 1)+\hat{k}(0 \times 1-41) \\ &=-7 \hat{\imath}+3 \hat{\jmath}-4 \hat{k} \\\\ &|\vec{x} \times \vec{y}|=\sqrt{(-7)^{2}+3^{2}+(-4)^{2}} \\\\ &=\sqrt{49+9+16} \\\\ &\vec{a}=\sqrt{74} \end{aligned}$

Vector or Cross Product exercise 24.1 question 5

Answer : $\sqrt{504}$
Hint : To solve this we multiply $\vec{b}$by 2 as given then magnitude formula.
Given : $\vec{a}=4 \hat{\imath}+3 \hat{\jmath}+\hat{k}$
$\vec{b}=\hat{\imath}-2 \hat{k}$
Find $|2 \vec{b} \times \vec{a}|$
Solution :
We need to find unit vector of b ($\hat{b}$)
$\begin{aligned} &\hat{b}=\frac{\vec{b}}{|\vec{b}|} \\\\ &\hat{b}=\frac{(\hat{i}-2 \hat{k})}{\sqrt{1^{2}+(-2)^{2}}} \\\\ &\hat{b}=\frac{1}{\sqrt{5}}(\hat{i}-2 \hat{k}) \end{aligned}$
$\begin{array}{r} 2 \hat{b}=\frac{2}{\sqrt{5}}(\hat{\mathbf{i}}-2 \hat{k}) \\\\ 2 \vec{b} \times \vec{a}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \\\frac{2}{\sqrt{5}} & 0 & \frac{-4}{\sqrt{5}} \\\\ 4 & 3 & 1 \end{array}\right| \end{array}$
$\begin{aligned} &=\hat{\mathrm{i}}\left[0.1-3 \cdot\left(\frac{-4}{\sqrt{5}}\right)\right]-\hat{\mathrm{j}}\left[\left(\frac{2}{\sqrt{5}}\right) 1-(4)\left(\frac{-4}{\sqrt{5}}\right)\right]+\mathrm{k}\left[\left(\frac{2}{\sqrt{5}}\right) 3-(4)(0)\right] \\\\ &2 \vec{b} \times \vec{a}=\frac{12}{\sqrt{5}} \hat{\mathrm{i}}-\frac{18}{\sqrt{5}} \hat{\mathrm{j}}+\frac{6}{\sqrt{5}} \hat{\mathrm{k}} \end{aligned}$
$\begin{aligned} |2 \vec{b} \times \vec{a}| &=\sqrt{\left(\frac{12}{\sqrt{5}}\right)^{2}+\left(\frac{-18}{\sqrt{5}}\right)^{2}+\left(\frac{6}{\sqrt{5}}\right)^{2}} \\\\ &=\sqrt{\frac{144}{5}+\frac{324}{5}+\frac{36}{5}} \\\\ |2 \vec{b} \times \vec{a}| &=\sqrt{\frac{504}{5}} \end{aligned}$

Vector or Cross Product exercise 24.1 question 6

Answer : $-25 \hat{\imath}+35 \hat{\jmath}-55 \hat{k}$
Hint : To solve this equation we put value in $(\vec{a}+2 \vec{b}) \text { and }(2 \vec{a}-\vec{b})$
Given :$\vec{a}=3 \hat{\imath}-\hat{\jmath}-2 \hat{k}$
$\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}$
Find $(\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})$
Solution :
$\begin{aligned} &(\vec{a}+2 \vec{b})=(3 \hat{\imath}-\hat{\jmath}-2 \hat{k}+2(2 \hat{\imath}+3 \hat{j}+\hat{k})) \\\\ &=3 \hat{i}+4 \hat{i}-\hat{j}+6 \hat{j}-2 \hat{k}+2 \hat{k} \\\\ &=7 \hat{\imath}+5 \hat{\jmath} \end{aligned}$...............(1)
$\begin{aligned} &(2 \vec{a}-\vec{b})=2(3 \hat{\imath}-\hat{\jmath}-2 \hat{k})-(2 \hat{\imath}+3 \hat{\jmath}+\hat{k}) \\\\ &=6 \hat{\imath}-2 \hat{\imath}-2 \hat{\jmath}-3 \hat{\jmath}-4 \hat{k}-\hat{k} \\\\ &=4 \hat{\imath}-5 \hat{\jmath}-5 \hat{k} \end{aligned}$...............(2)
$\begin{aligned} &(\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 7 & 5 & 0 \\ 4 & -5 & -5 \end{array}\right| \\\\ &=\hat{\imath}(-25-0)-\hat{\jmath}(-35-0)+\hat{k}(-35-20) \\\\ &=-25 \hat{\imath}+35 \hat{\jmath}-55 \hat{k} \end{aligned}$

Vector or Cross Product exercise 24.1 question 7 (i)

Answer : $42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k}$
Hint : To solve this equation we solve $\vec{a} \times \vec{b} \text { then }|\vec{a} \times \vec{b}|$
Given : $2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \text { and } 3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}$ magnitude=49
Solution : $\frac{\vec{a} \times \vec{b}}{(\vec{a} \times \vec{b})}=49$
$\begin{aligned} &\vec{a}=2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \\\\ &\vec{b}=3 \hat{\imath}-6 \hat{\jmath}+2 \hat{b} \\\\ &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{array}\right| \end{aligned}$
$\begin{aligned} &=\hat{\imath}(42)-\hat{\jmath}(-14)+\hat{k}(-2) \\\\ &=42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k} \\\\ &=7(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \end{aligned}$
$\begin{aligned} &|\vec{a} \times \vec{b}|=7 \sqrt{6^{2}+2^{2}+(-3)^{2}} \\\\ &=7 \sqrt{36+4+9} \\\\ &=7 \sqrt{49} \quad=49 \\\\ &42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k} \end{aligned}$

Vector or Cross Product exercise 24.1 question 7 (ii)

Answer : $2 \hat{\imath}-2 \hat{\jmath}+\hat{k}$
Hint : To solve this equation we use unit vector $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$ formula
Given :$\vec{a}=3 \hat{\imath}+\hat{\jmath}-4 \hat{k} \vec{b}=6 \hat{\imath}+5 \hat{\jmath}-2 \hat{k}$ magnitude = 3
Solution :$3 \times \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{j} & \hat{k} \\ 3 & 1 & -4 \\ 6 & 5 & -2 \end{array}\right|$
$\begin{aligned} &=\hat{\imath}(1 \times-2-(5 \times-4))-\hat{\jmath}(3 \times-2-(-4) \times 6)+\hat{k}(3 \times 5-1 \times 6) \\\\ &=18 \hat{\imath}-18 \hat{\jmath}+9 \hat{k} \\\\ &=9(2 \hat{\imath}-2 \hat{\jmath}+\hat{k}) \end{aligned}$
$\begin{aligned} &|\vec{a} \times \vec{b}|=9 \sqrt{2^{2}+(-2)^{2}+1^{2}} \\\\ &=9 \sqrt{4+4+1} \\\\ &=9 \sqrt{9} \\\\ &=9 \times 3 \end{aligned}$
$\begin{aligned} &=27 \\\\ &3 \times \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{3 \times 9(2 \hat{i}-2 \hat{j}+\hat{k})}{27} \\\\ &=2 \hat{\imath}-2 \hat{j}+\hat{k} \end{aligned}$

Vector or Cross exercise 24.1 question 8 (i)

Answer : 6 square units
Hint : To solve this we use area of parallelogram
Given : $2 \hat{\imath} \text { and } 3 \hat{\jmath}$
Solution : Area of parallelogram
$\begin{aligned} &=|\vec{a} \times \vec{b}| \\\\ &=|2 \hat{\imath} \times 3 \hat{\jmath}| \\\\ \end{aligned}$
$\begin{aligned} &=|6 \hat{k}| \\\\ &=6 \; s q \cdot \text { units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 8 (ii)

Answer : $3 \sqrt{3} \text { sq. units }$
Hint : To solve this , we use area of parallelogram.
Given :$2 \hat{\imath}+\hat{\jmath}+3 \hat{k} ; \hat{\imath}-\hat{\jmath}$
Solution : Area of parallelogram$=|\vec{a} \times \vec{b}|$
$\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right| \\\\ &=\hat{i}(1 \times 0-3(-1))-\hat{j}(2 \times 0-3 \times 1)+\hat{k}(2 \times 1-(-1) \times 1) \\\\ &=3 \hat{\imath}+3 \hat{j}+3 \hat{k} \end{aligned}$
Area of parallelogram
$\begin{aligned} &=\sqrt{3^{2}+3^{2}+3^{2}} \\\\ &=\sqrt{9+9+9} \\\\ &=\sqrt{27} \\\\ &=3 \sqrt{3} \text { sq. units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 8 (iii)

Answer : $10 \sqrt{3}\; s q . \text { units }$
Hint : To solve this equation we use area of parallelogram
Given : $3 \hat{\imath}+\hat{\jmath}-2 \hat{k} \quad \text { and } \hat{\imath}-3 \hat{\jmath}+4 \hat{k}$
Solution : Area of parallelogram $=|\vec{a} \times \vec{b}|$
$\begin{aligned} &\vec{a}=3 \hat{\imath}+\hat{\jmath}-2 \hat{k} \\\\ &\vec{b}=\hat{\imath}-3 \hat{j}+4 \hat{k} \end{aligned}$
$\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{array}\right| \\\\ &=\hat{\imath}(1 \times 4-(-2)(-3))-\hat{\jmath}(3 \times 4-2 \times-1)+\hat{k}(3 \times-3-1 \times 1) \\\\ &=-2 \hat{\imath}-14 \hat{\jmath}-10 \hat{k} \end{aligned}$
$\begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{(-2)^{2}+(-14)^{2}+(-10)^{2}} \\\\ &=\sqrt{4+196+100} \\\\ &=\sqrt{300} \\\\ &=10 \sqrt{3} \text { sq.units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 8 (iv)

Answer : $4 \sqrt{2} \text { sq. units }$
Hint : To solve this equation we use area of parallelogram
Given $\hat{\imath}-3 \hat{\jmath}+\hat{k} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}$
Solution :$\vec{a}=\hat{\imath}-3 \hat{\jmath}+\hat{k} \quad ; \quad \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k}$
Area of parallelogram $=|\vec{a} \times \vec{b}|$
$\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &\vec{a} \times \vec{b}=\hat{i}[-3.1-1.1]-\hat{j}[1.1-1.1]+\hat{k}[1.1-1 \cdot(-3)] \end{aligned}$
$\begin{aligned} &=\hat{\mathrm{i}}[-3-1]-\hat{\mathrm{j}}[1-1]+\hat{\mathrm{k}}[1+3] \\\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=-4 \hat{\mathbf{i}}+4 \hat{\mathrm{k}} \end{aligned}$
Area of parallelogram
$\begin{aligned} &=\sqrt{-4^{2}+4^{2}} \\\\ &=4 \sqrt{2} \end{aligned}$

Vector or Cross Product exercise 24.1 question 9 (i)

Answer : $\frac{15}{2} \text { sq.units }$
Hint : To solve this we use area of parallelogram formula
Given : Area of parallelogram $=\frac{1}{2}\left(d_{1} \times d_{2}\right)$
Solution : $d_{1}=-2 \hat{\imath}+\hat{\jmath}-2 \hat{k}$
$\begin{aligned} &d_{2}=4 \hat{\imath}-\hat{\jmath}-3 \hat{k} \\\\ &d_{1} \times d_{2}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -2 & 1 & -2 \\ 4 & -1 & -3 \end{array}\right| \end{aligned}$
$\begin{aligned} &=\hat{\imath}(1 \times-3-(-2)(-1))-\hat{j}(-2 \times-3-4 \times 2)+\hat{k}(-2 \times-1-4 \times-1) \\\\ &=-5 \mathrm{i}-14 \mathrm{j}-2 \mathrm{k} \end{aligned}$
$\begin{aligned} &A=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ &=\frac{1}{2} \sqrt{(-5)^{2}+14^{2}+(-2)^{2}} \\\\ &=\frac{1}{2} \sqrt{25+196+4} \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \sqrt{225} \\\\ &=\frac{15}{2} \text { sq.units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 9 (ii)

Answer : $\frac{\sqrt{6}}{2} \text { sq.units }$
Hint : To solve this we use area of parallelogram formula
Given : $2 \hat{\imath}+\hat{k} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}$
Solution : Area of parallelogram $=\frac{1}{2}\left(d_{1} \times d_{2}\right)$
$\begin{gathered} d_{1}=2 \hat{\imath}+\hat{k} \\\\ d_{2}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{gathered}$
$\begin{aligned} &d_{1} \times d_{2}\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(0 \times 1-1 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2) \\\\ &=-\hat{\imath}-\hat{\jmath}+2 \hat{k} \end{aligned}$
$\begin{aligned} &\frac{1}{2}\left|d_{1} \times d_{2}\right|=\frac{1}{2} \sqrt{(-1)^{2}+(-1)^{2}+2^{2}} \\\\ &=\frac{1}{2} \sqrt{1+1+4} \\\\ &=\frac{\sqrt{6}}{2} \text { sq.units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 9 (iii)

Answer : $\frac{\sqrt{26}}{2} \text { sq.units }$
Hint : To solve this we use area of parallelogram formula
Given : $3 \hat{\imath}+4 \hat{\jmath} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}$
Solution :Area of parallelogram $=\frac{1}{2}\left(d_{1} \times d_{2}\right)$
$\begin{gathered} d_{1}=3 \hat{\imath}+4 \hat{\jmath} \\ d_{2}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{gathered}$
$\begin{aligned} &d_{1} \times d_{2}=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(4 \times 1-0 \times 1)-\hat{j}(3 \times 1-0 \times 1)+\hat{k}(3 \times 1-4 \times 1) \\\\ &=4 \hat{\imath}-3 \hat{\jmath}-1 \hat{k} \end{aligned}$
$\begin{aligned} &A \quad=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ &=\frac{1}{2} \sqrt{(-4)^{2}+(-3)^{2}+(-1)^{2}} \\\\ &=\frac{1}{2} \sqrt{16+9+1} \\\\ &=\frac{\sqrt{26}}{2} \text { sq.units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 9 (iv)

Answer : $\frac{49}{2} \text { sq.units }$
Hint : To solve this we use area of parallelogram formula
Given : $d_{1}=2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k} \quad, d_{2}=3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}$
Solution : Area of parallelogram $=\frac{1}{2}\left(d_{1} \times d_{2}\right)$
$\begin{aligned} &d_{1} \times d_{2}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{array}\right| \\\\ \quad &=\hat{\imath}(3 \times 2-(-6 \times 6))-\hat{\jmath}(2 \times 2-3 \times 6)+\hat{k}(2 \times-6-3 \times 3) \\\\ &\quad=42 \hat{\imath}+14 \hat{\jmath}-21 \hat{k} \end{aligned}$
$\begin{gathered} A=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ =\frac{1}{2} \sqrt{42^{2}+14^{2}+-21^{2}} \\\\ =\frac{1}{2} \sqrt{144+196+9} \end{gathered}$
$\begin{aligned} &=\frac{1}{2} \sqrt{349} \\\\ &=\frac{49}{2} \text { sq.units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 10

Answer : not equal
Hint : To solve this w use determinant method
Given :
$\begin{aligned} &\vec{a}=2 \hat{\imath}+5 \hat{\jmath}-7 \hat{k} \\ &\vec{b}=3 \hat{i}+4 \hat{\jmath}+\hat{k} \\ &\vec{c}=\hat{\imath}-2 \hat{\jmath}-3 \hat{k} \end{aligned}$
Solution : $\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{j} & \hat{k} \\ 2 & 5 & -7 \\ -3 & 4 & 1 \end{array}\right|$
$\begin{aligned} &\vec{a}=33 \hat{\imath}+19 \hat{\jmath}+23 \hat{k} \\\\ &\vec{b} \times \vec{c}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -3 & 4 & 1 \\ 1 & -2 & -3 \end{array}\right| \\\\ &\vec{l}=-10 \hat{\imath}-8 \hat{\jmath}+2 \hat{k} \end{aligned}$
$\begin{aligned} &\vec{d} \times \vec{c}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 33 & 19 & 23 \\ 1 & -2 & -3 \end{array}\right| \\\\ &(\vec{a} \times \vec{b}) \times \vec{c}=-11 \hat{\imath}+122 \hat{\jmath}-85 \hat{k} \\\\ &\vec{a} \times \vec{l}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 5 & -7 \\ -10 & -8 & 2 \end{array}\right| \end{aligned}$
$\begin{aligned} &\vec{a} \times(\vec{b} \times \vec{c})=-46 \hat{\imath}+66 \hat{\jmath}+34 \hat{k} \\\\ &(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a}(\vec{b} \times \vec{c}) \end{aligned}$

Vector or Cross Product exercise 24.1 question 11

Answer : 6

Hint : To solve this we use $\vec{a}\; \&\; \vec{b}$ formula
Given : $|\vec{a}|=2 ;|\vec{b}|=5 \text { and }(\vec{a} \times \vec{b})=8 \text { find } \vec{a} \cdot \vec{b}$
Solution : $|\vec{a} \times \vec{b}|=|\vec{a}| \cdot|\vec{b}| \sin \theta$
$\begin{aligned} &8=2 \times 5 \sin \theta \\\\ &8=10 \sin \theta \\\\ &\sin \theta=\frac{8}{10}=\frac{4}{5} \end{aligned}$
$\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &=2 \times 5 \times \frac{3}{5} \\\\ &=6 \end{aligned}$

Vector or Cross Product exercise 24.1 question 12

Answer : These are unit vectors as well as perpendicular
Hint : To solve this , we do magnitude of one by one
Given :
$\begin{aligned} &\vec{a}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \\\\ &\vec{b}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}) \\\\ &\vec{c}=\frac{1}{7}(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \end{aligned}$

Solution : $|\vec{a}|=\frac{1}{7} \sqrt{2^{2}+3^{2}+6^{2}}$

$\begin{aligned} &=\frac{1}{7} \sqrt{49}=>\frac{7}{7}=1 \\\\ &|\vec{b}|=\frac{1}{7} \sqrt{49}=1 \\\\ &|\vec{c}|=\frac{1}{7} \sqrt{36+4+9} \end{aligned}$
$\begin{aligned} &=\frac{1}{7} \sqrt{49}=\frac{7}{7}=1 \\\\ &\vec{a} \cdot \vec{b}=0 \\\\ &\vec{b} \cdot \vec{c}=0 \end{aligned}$
$\begin{aligned} &\vec{a} \cdot \vec{b}=\frac{1}{49}[6-18+12] \\\\ &\vec{b} \cdot \vec{c}=\frac{1}{49}[18-12-6] \end{aligned}$

Vector or Cross Product exercise 24.1 question 13

Answer : 25
Hint : To solve this formula $\vec{a} \cdot \vec{b} \text { and }|\vec{a} \times \vec{b}|$
Given : $|\vec{a}|=13 \quad ;|\vec{b}|=5$
$\vec{a} \cdot \vec{b}=60$
Solution : $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$
$\begin{aligned} &60=(13)(5) \cos \theta \\\\ &\cos \theta=\frac{60}{13 \times 5}=\frac{12}{13} \end{aligned}$
$\begin{aligned} &\cos ^{2} \theta+\sin ^{2} \theta=1 \\\\ &\sin \theta=\sqrt{1+\cos ^{2} \theta} \\\\ &=\sqrt{1-\left(\frac{12}{13}\right)^{2}} \end{aligned}$
$\begin{aligned} &=\sqrt{1-\frac{144}{169}} \\\\ &=\sqrt{\frac{169-144}{169}} \\\\ &=\sqrt{\frac{25}{169}} \end{aligned}$
$\begin{aligned} &=\frac{5}{13} \\\\ &|\vec{a} \times \vec{b}|=|| \vec{a}|| \vec{b}|\sin \theta| \\\\ &=13 \times 5 \times \frac{5}{13} \\\\ &=25 \end{aligned}$

Vector or Cross Product exercise 24.1 question 14

Answer :$\frac{\pi }{4}$
Hint : To solve this equation we use aband $|\vec{a}||\vec{b}| \text { and }|\vec{a} \times \vec{b}|$
Given : $|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}$
Solution :
$1)\vec{a} \times \vec{b}|=|a||b| \cos \theta\\$
$2) \vec{a} \cdot \vec{b}|=|a||b| \sin \theta$
$\Rightarrow|\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| \sin \theta$
$\begin{aligned} &\Rightarrow \frac{\sin \theta}{\cos \theta}=1 \\\\ &\Rightarrow \tan \theta=1 \\\\ &\Rightarrow \theta=\tan ^{-1} 1 \\\\ &\theta=\frac{\pi}{4} \end{aligned}$

Vector or Cross Product exercise 24.1 question 15

Answer : $\vec{a}+\vec{c}=m \vec{b}$
Hint : Here m is the constant team
Given :$\vec{a} \times \vec{b}=\vec{b} \times \vec{c}$
$\vec{a} \times \vec{b}-\vec{b} \times \vec{c}=0$
Solution :
$\begin{aligned} &\vec{b} \times \vec{c}=-\vec{c} \times \vec{b} \text { (substitute) } \\\\ &\vec{a} \times \vec{b}+\vec{c} \times \vec{b}=0 \\\\ &(\vec{a}+\vec{c}) \times \vec{b}=0 \\\\ &\vec{a}+\vec{c}=m \vec{b} \end{aligned}$

Vector or Cross Product exercise 24.1 question 16

Answer : $\frac{\pi }{6}$
Hint : To solve this we use $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$ formula
Given : $\vec{a} \times \vec{b}=3 \hat{\imath}+2 \hat{\jmath}+6 \hat{k}$
Solution : $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$
$\begin{gathered} \sqrt{3^{2}+2^{2}+6^{2}}=2 \times 7 \sin \theta \\\\ \sqrt{9+4+12}=14 \sin \theta \\\\ 7 =14 \sin \theta \end{gathered}$
$\begin{aligned} \sin \theta &=\frac{1}{2} \\\\ \theta &=\frac{\pi}{6} \end{aligned}$

Vector or Cross Product exercise 24.1 question 17

Answer : $\vec{a} \cdot \vec{b}=0 ; \vec{a} \times \vec{b}=0$
Hint : To solution we know $\vec{a}, \vec{b}$ are parallel
Given :$\vec{a} \times \vec{b}=0$
Solution : 1) $\vec{a}$ and $\vec{b}$ are parallel
Or
$\begin{aligned} &|\vec{a}|=0 \text { or }|\vec{b}|=0 \\\\ &\vec{a} \cdot \vec{b}=0 \end{aligned}$

$\vec{a}$ & $\vec{b}$ are perpendicular
$|\vec{a}|=0 \text { or }|\vec{b}|=0$

Either $|\vec{a}|=0 \text { or }|\vec{b}|=0$ or both are zero if $\vec{a} \cdot \vec{b}=0\; \& \; \vec{a} \times \vec{b}=0$

Vector or Cross Product exercise 24.1 question 18

Answer:

Hint : To solve this equation we use $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$
Given : $\vec{a} \times \vec{b}=\vec{c} \text { and } \vec{b} * \vec{c}=\vec{a} \text { and } \vec{c} * \vec{a}=\vec{b}$
Solution : $\vec{a} \times \vec{b}=\vec{c}$
$\begin{aligned} &\Rightarrow \vec{a} \times \vec{b}=\vec{c} \\\\ &\Rightarrow|\vec{a}||\vec{b}| \sin \theta=\vec{c} \\\\ &\theta=\sin ^{-1}(1) \\\\ &\theta=\frac{\Pi}{2} \; 0 r \; 90^{\circ} \end{aligned}$

Similarly can be prove for others

$\vec{b} \times \vec{c}=\vec{a}$ this means perpendicular $(\vec{b} \times \vec{c})$
$\vec{c} \times \vec{a}=\vec{b}$ this means perpendicular $(\vec{c} \times \vec{a})$
This are together $\vec{a}, \vec{b} \text { and } \vec{c}$ form orthonormal trial

Vector or Cross Product exercise 24.1 question 19

Answer : $\frac{1}{\sqrt{165}}(10 \hat{\imath}+7 \hat{\jmath}-6 \hat{k})$
Hint : To solve this equation , we use determination method
Given : $A=(3,-1,2)$

$\begin{aligned} &B=(1,-1,-3) \\\\ &C=(4,-3,1) \end{aligned}$
Solution : $\overrightarrow{A B}=-2 \hat{\imath}-5 \hat{k}$
$\begin{aligned} &\overrightarrow{A C}=\hat{\imath}-2 \hat{\jmath}-\hat{k} \\\\ &\vec{C}=x \overrightarrow{A B}+y \overrightarrow{A C} \end{aligned}$
If $\vec{a}$ is perpendicular to $\vec{c}$
$\vec{d}$ is perpendicular to $\overrightarrow{AB}$ ,$\vec{d}$ perpendicular to $\overrightarrow{AC}$
$\begin{aligned} &\vec{d}=\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{array}\right| \\\\ &=-10 \hat{\imath}-7 \hat{\jmath}+4 \hat{k} \\\\ &\hat{d}=\frac{\vec{d}}{|\vec{d}|}=\frac{-10 \hat{\imath}-7 \hat{\jmath}+4 \hat{k}}{\sqrt{165}} \end{aligned}$

Vector or Cross Product exercise 24.1 question 20

Answer:
To have $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
Hint:
To solve this we use $|\vec{a} \times \vec{b}| \text { and }|\vec{a}||\vec{b}|$ formula
Given:
Solution : $\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B}=0$
$\begin{aligned} &\overrightarrow{A B} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{A B} \times 0 \\\\ &a c(\sin B-\sin A)=0 \end{aligned}$
Divide by abc
$\begin{aligned} &\frac{\sin B}{b}-\frac{\sin A}{a}=0 \\\\ &\frac{\sin B}{b}=\frac{\sin A}{a} \end{aligned}$ .......(i)
$\begin{aligned} &\overrightarrow{B C} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{B C} \times 0 \\\\ &\frac{\sin B}{b}=\frac{\sin C}{c} \end{aligned}$.......(ii)
$\begin{aligned} &\overrightarrow{C A} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{B C} \times 0 \\\\ &\frac{\sin C}{c}=\frac{\sin A}{a} \end{aligned}$.......(iii)
From (i),(ii) and (iii)

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Vector or Cross Product exercise 24.1 question 21

Answer: $\hat{i}+11 \hat{j}+7 \hat{k}$
Hint:
To solve this we use determinant method
Given:
$\begin{aligned} &\vec{c}=\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{array}\right| \\\\ &=\hat{i}+11 \hat{j}+7 \hat{k} \end{aligned}$
Solution:
$\begin{aligned} &\hat{i}+11 \hat{j}+7 \hat{k} \\\\ &\vec{c} \cdot \vec{a}=1-22+21=0 \\\\ &\vec{c} \text { is } \perp \text { to } \vec{a} \end{aligned}$

Vector or Cross Product exercise 24.1 question 22

Answer: $\frac{3}{4}$ Square units
Hint: use concept.
Given:
$A=\frac{1}{2}|\vec{a} \times \vec{b}|$
Solution:
$\begin{aligned} &A=\frac{1}{2}|\vec{p}+2 \vec{q}| \times|2 \vec{p}+\vec{q}| \\\\ &=\frac{1}{2}|0| \times \vec{p} \times \vec{q}+4|\vec{p}+\vec{q}|+0 \\\\ &\{\vec{p} \times \vec{p}=0\} \end{aligned}$
$\begin{aligned} &=\frac{1}{2}|\vec{q} \times \vec{p}+4 \vec{q} \times \vec{p}| \\\\ &=\frac{1}{2} \times 3|\vec{q} \times \vec{p}| \end{aligned}$
$\begin{aligned} &=\frac{3}{2}|\vec{p} \| \vec{q}| \sin \theta \\\\ &=\frac{3}{2} \cdot \frac{1}{2} \end{aligned}$
$=\frac{3}{4}$ Square units

Vector or Cross Product exercise 24.1 question 23

Answer:
L.H.S = R.H.S
Hint:
To solve this we use formula
Given:
$|\vec{a} \times \vec{b}|^{2}=\left|\begin{array}{ll} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{array}\right|$
Solution:
$RHS \\\\\begin{aligned} &\left|\begin{array}{ll} \left|\vec{a}^{2}\right| & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{array}\right| \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{a}) \end{aligned}$
$=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \quad[\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}]$
$\begin{aligned} &=|\vec{a}|^{2}|\vec{b}|^{2}-(|\vec{a}||\vec{b}| \cos \theta)^{2} \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right) \end{aligned}$
$\begin{aligned} &=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \\\\ &=|\vec{a} \times \vec{b}|^{2} \quad[\because \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta] \end{aligned}$

Vector or Cross Product exercise 24.1 question 24

Answer: $|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b}) \tan \theta$
Hint:
Given: $|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b}) \tan \theta$
Solution:
$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$ .........(1)
Also,
$|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}| \cos \theta$ ............(2)
Dividing equation (1) and (2)
$\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=\frac{|\vec{a}| \vec{b} \mid \sin \theta}{|\vec{a}||\vec{b}| \cos \theta}$
$\begin{aligned} &\frac{|\vec{a} \times \vec{b}|}{(\vec{a} \cdot \vec{b})}=\tan \theta \\\\ &|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b}) \tan \theta \end{aligned}$

Vector or Cross Product exercise 24.1 question 25

Answer: 7
Hint:
To solve this we use $|\vec{a} \times \vec{b}|$ formula
Given:
$\begin{aligned} &|\vec{a}|=\sqrt{26} ;|\vec{b}|=7 \\\\ &(\vec{a} \times \vec{b})=35 \end{aligned}$
Solution:
$\begin{aligned} &|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \\\\ &|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right) \\\\ &|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \end{aligned}$
$\begin{aligned} &(35)^{2}=(\sqrt{26})^{2}(7)^{2}-(\vec{a} \cdot \vec{b})^{2} \\\\ &|\vec{a} \cdot \vec{b}|=\sqrt{49} \\\\ &\vec{a} \cdot \vec{b}=7\ \end{aligned}$

Vector or Cross Product exercise 24.1 question 26

Answer:
$3 \sqrt{5}$square units
Hint:
To solve this we use area of triangle formula
Given:
$\begin{aligned} &\overrightarrow{O A}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\ &\overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k} \end{aligned}$
Solution:
$\begin{aligned} &A=\frac{1}{2}|\overrightarrow{O A} \times \overrightarrow{O B}| \\\\ &=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &=\frac{1}{2}|8 \hat{i}-10 \hat{j}+4 \hat{k}| \\\\ &=\frac{1}{2} \sqrt{64+100+16} \end{aligned}$
$=\frac{\sqrt{180}}{2}=\sqrt{45}\\\\$
$=3 \sqrt{5}$ square units

Vector or Cross Product exercise 24.1 question 27 (i)

Answer:
$\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k})$
Hint: To solve this we use determinant method
Given:
$\begin{aligned} &\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k} \\\\ &\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k} \\\\ &\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k} \\\\ &\vec{c} \cdot \vec{d}=15 \end{aligned}$
Solution:
D is perpendicular to a and b both. Hence, parallel to a*b
$\begin{aligned} &\vec{a} \times \vec{b}=\vec{c} \\\\ &\vec{a} \times \vec{b}=\vec{d} \\\\ &\vec{d}=\lambda\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{array}\right| \end{aligned}$
$\begin{aligned} &\vec{d}=\lambda(32 \hat{i}-\hat{j}-14 \hat{k}) \\\\ &\vec{c} \cdot \vec{d}=15 \\\\ &\lambda(2 \hat{i}-\hat{j}+4 \hat{k})(32 \hat{i}-\hat{j}-14 \hat{k})=15 \end{aligned}$
$\begin{aligned} &\lambda(64+1-56)=15 \\\\ &9 \lambda=15 \\\\ &\lambda=\frac{5}{3} \\\\ &\vec{d}=\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k}) \end{aligned}$

Vector or Cross Product exercise 24.1 question 27 (ii)

Answer:
$\vec{a}=7(\hat{i}-\hat{j}-\hat{k})$
Hint:
To solve this question we suppose term in terms of x,y,z.
Given:
$\begin{aligned} &\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k} \\\\ &\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k} \\\\ &\vec{c}=3 \hat{i}+\hat{j}-\hat{k} \\\\ &\vec{d} \cdot \vec{c}=21 \end{aligned}$
Solution: Let $\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}$
Now, $\vec{d} \cdot \vec{c}=0$
$\begin{aligned} &(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}+\hat{j}-\hat{k})=0 \\\\ &3 x+y-z=0 \end{aligned}$.......(i)
Now, $\vec{d} \cdot \vec{b}=0$
$\begin{aligned} &(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-4 \hat{j}+5 \hat{k})=0 \\\\ &x-4 y+5 z=0 \end{aligned}$....(ii)
Now,
$\begin{aligned} &\vec{d} \cdot \vec{a}=0 \\\\ &(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(4 \hat{i}+5 \hat{j}-\hat{k})=0 \\\\ &4 x+5 y-z=0 \end{aligned}$....(iii)
Solving (2) and (3)
$\begin{aligned} &\frac{x}{25-4}=\frac{y}{-1-20}=\frac{z}{-16-5}=k \\\\ &\frac{x}{-21}=\frac{y}{-21}=\frac{z}{-21}=k \end{aligned}$
$\begin{aligned} &\frac{x}{1}=\frac{y}{-1}=\frac{z}{-1}=k \\\\ &x=1 ; y=-1 ; z=-k \end{aligned}$
Putting value in (i)
$\begin{aligned} &3 k-k+k=21 \\\\ &k=7 \\\\ &x=7 ; y=-7 ; z=-7 \\\\ &\vec{a}=7(\hat{i}-\hat{j}-\hat{k}) \end{aligned}$

Vector or Cross Product exercise 24.1 question 28

Answer:
$\frac{1}{3}(2 \hat{i}-2 \hat{j}-\hat{k})$
Hint:
To solve this we use determinant method
Given:
$\begin{aligned} &\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k} \\\\ &\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \end{aligned}$
Solution:
$\begin{aligned} &\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k} \\\\ &\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \end{aligned}$
Let
$\begin{aligned} &\vec{d}=\vec{a}+\vec{b} \\\\ &\vec{d}=4 \hat{i}+4 \hat{j}+0 \hat{k} \end{aligned}$
And
$\begin{aligned} &\vec{e}=\vec{a}-\vec{b} \\\\ &=2 \hat{i}+0 \hat{j}+4 \hat{k} \end{aligned}$
Let $\vec{f}$ be any vector perpendicular to both $\vec{d} \; \& \; \vec{e}$, hence parallel to $\vec{d} \times \vec{e}$
$\begin{aligned} &\therefore \vec{f}=\vec{d} \times \vec{e} \\\\ &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{array}\right| \\\\ &=16 \hat{i}-16 \hat{j}-8 \hat{k} \end{aligned}$
$\begin{aligned} &=\frac{16\hat {i}-16 \hat{j}-8 \hat {k}}{\sqrt{(16)^{2}+(-16)^{2}+(-8)^{2}}} \\\\ &=\frac{16}{24} \hat{i}-\frac{16}{24} \hat{j}-\frac{8}{24} \hat{k} \\\\ &=\frac{1}{3}(2 \hat{i}-2 \hat{j}-\hat{k}) \end{aligned}$

Vector or Cross Product exercise 24.1 question 29

Answer:
$\frac{\sqrt{61}}{2} \text { square units }$
Hint:
To solve this we use area of triangle ABC
Given:
$\begin{aligned} &A(2,3,5) \\ &B(3,5,8) \\ &C(2,7,8) \end{aligned}$
$\text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|$
Solution:
$\begin{aligned} &\overrightarrow{A B}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\ &\overrightarrow{B C}=-\hat{i}+2 \hat{j} \\\\ &\overrightarrow{A B} \times \overrightarrow{B C}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 2 & 0 \end{array}\right| \end{aligned}$
$\begin{aligned} &=\hat{i}(0-6)-\hat{j}(0+3)+\hat{k}(2+2) \\\\ &=-6 \hat{i}-3 \hat{j}+4 \hat{k} \\\\ &|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{(-6)^{2}+(-3)^{2}+(4)^{2}} \end{aligned}$
$\begin{aligned} &|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{36+9+16} \\\\ &|\overrightarrow{A B} \times \overrightarrow{B C}|=\sqrt{61}\\\\ \end{aligned}$
$Area \; of \; \Delta \mathrm{ABC}=\frac{1}{2}|\sqrt{61}|$
$=\frac{\sqrt{61}}{2} square\; units$

Vector or Cross Product exercise 24.1 question 30

Answer:
$\frac{\sqrt{21}}{2}$ square units
Hint:
To solve this we use area of triangle
Given:
$\begin{aligned} &\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k} \\\\ &\vec{b}=-\hat{i}+\hat{k} \\\\ &\vec{c}=2 \hat{j}-\hat{k} \end{aligned}$
$\begin{aligned} &\vec{a}+\vec{b}=\hat{i}-3 \hat{j}+2 \hat{k} \\\\ &\vec{b}+\vec{c}=-\hat{i}+2 \hat{j} \end{aligned}$
Solution:
$\begin{aligned} &A=\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \\\\ &A=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{array}\right| \end{aligned}$
$\begin{aligned} &=\frac{1}{2}|-4 \hat{i}-2 \hat{j}-\hat{k}| \\\\ &=\frac{1}{2} \sqrt{16+4+1} \\\\ &=\frac{\sqrt{21}}{2} \text { square units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 31

Answer:
$11 \sqrt{5} \text { square units }$
Hint:
To solve this we use $\vec{c}=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}$
Given:
$\begin{aligned} &\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k} \\\\ &\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k} \end{aligned}$
Solution:
$\begin{aligned} &\vec{c}=\vec{a}+\vec{b}=3 \hat{i}-6 \hat{j}+2 \hat{k} \\\\ &\vec{c}=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|} \\\\ &=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k} \end{aligned}$
$\begin{aligned} &A=|\vec{a} \times \vec{b}|=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{array}\right| \\\\ &=|22 \hat{i}+11 \hat{j}+0 \hat{k}| \\\\ &=11|2 \hat{i}+\hat{j}| \end{aligned}$
$\begin{aligned} &=11 \sqrt{(2)^{2}+(1)^{2}} \\\\ &=11 \sqrt{5} \end{aligned}$

Vector or Cross Product exercise 24.1 question 32

Answer:
No, take any two collinear vectors
Hint:
To solve this we let
$\hat{i}+\hat{j}+\hat{k}=\overrightarrow{0} ; 2 \hat{i}+2 \hat{j}+2 \hat{k}=\overrightarrow{0}$
Given: $\vec{a}=0 ; \vec{b}=0 \text { then } \vec{a} \times \vec{b}=0$
Solution:
Statement:
If either $\vec{a}=\vec{b} \text { or } \vec{b}=\overrightarrow{0} \Rightarrow \vec{a} \times \vec{b}=0$
Converse: $\vec{a}=0 ; \vec{b}=0 \text { then } \vec{a} \times \vec{b}=0$
Let
$\begin{aligned} &\vec{a}=\hat{i}+\hat{j}+\hat{k}=(1,1,1) \\\\ &\vec{b}=2 \hat{i}+2 \hat{j}+2 \hat{k}=(2,2,2) \end{aligned}$
$\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}\right| \\\\ &=\hat{i}(2-2)+\hat{j}(2-2)+\hat{k}(2-2) \\\\ &=\vec{o} \end{aligned}$
But $\vec{a} \neq 0 \text { and } \vec{b} \neq 0$

Vector or Cross Product exercise 24.1 question 33

Answer:
Proved
Hint:
To solve this we use determinant method
Given:
$\begin{aligned} &a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k} \\\\ &b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k} \\\\ &c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k} \end{aligned}$
Solution:
$\vec{b}+\vec{c}=\left(b_{1}+c_{1}\right) \hat{i}+\left(b_{2}+c_{2}\right) \hat{j}+\left(b_{3}+c_{3}\right) \hat{k}$
$\vec{a} \times(\vec{b}+\vec{c})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3} \end{array}\right|$
$=\hat{i}\left[a_{2}\left(b_{3}+c_{3}\right)-a_{3}\left(b_{2}+c_{2}\right)\right]-\hat{j}\left[a_{1}\left(b_{3}+c_{3}\right)-a_{3}\left(b_{1}+c_{1}\right)\right]+\hat{k}\left[a_{1}\left(b_{2}+c_{2}\right)-a_{2}\left(b_{1}+c_{1}\right)\right]$
$\begin{aligned} &=\hat{i}\left(a_{2} b_{3}+a_{3} b_{2}\right)-\hat{j}\left(a_{1} b_{3}+a_{3} b_{1}\right)+\hat{k}\left(a_{1} b_{2}+a_{2} b_{1}\right)+\hat{i}\left(a_{2} c_{3}+a_{3} c_{2}\right)-\hat{j}\left(a_{1} c_{3}+a_{3} c_{1}\right)+\hat{k}\left(a_{1} c_{2}+a_{2} c_{1}\right) \\\\ &\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{c})+(\vec{a} \times \vec{b}) \end{aligned}$

Vector or Cross Product exercise 24.1 question 34

Answer:
$\sqrt{61}$ Square units
Hint:
To solve this we use area of triangle
Given:
$A(1,1,2) ; B(2,3,5) ; C(1,5,5)$
Solution:
$\begin{aligned} &D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{B C}) \\\\ &\Rightarrow \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=(2 \hat{i}+3 \hat{j}+5 \hat{k})-(\hat{i}+\hat{j}+2 \hat{k}) \\\\ &=\hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}$
$\begin{aligned} &\Rightarrow \overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \\\\ &=(\hat{i}+5 \hat{j}+5 \hat{k})-(2 \hat{i}+3 \hat{j}+5 \hat{k}) \\\\ &=-\hat{i}+2 \hat{j} \end{aligned}$
$\begin{aligned} &D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{B C}) \\\\ &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 2 & 0 \end{array}\right| \end{aligned}$
$\begin{aligned} &=\hat{i}(0-6)+\hat{j}(0-(-3))+\hat{k}(2-(-2)) \\\\ &=-6 \hat{i}-3 \hat{j}+4 \hat{k} \end{aligned}$
$\begin{aligned} &=(\overrightarrow{A B} \times \overrightarrow{B C})=\sqrt{(-6)^{2}+(-3)^{2}+(4)^{2}} \\\\ &=\sqrt{36+9+6} \\\\ &=\sqrt{61} \text { square units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 35

Answer:
$\frac{1}{2} \sqrt{274}$ Square units
Hint:
To solve this we use area of triangle
Given:
$A(1,2,3) ; B(2,-1,4) ; C(4,5,-1)$
Solution: $D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{A C})$
$\begin{aligned} &\Rightarrow \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=(2 \hat{i}-\hat{j}+4 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \\\\ &=\hat{i}-3 \hat{j}+\hat{k} \end{aligned}$
$\begin{aligned} &\Rightarrow \overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \\\\ &=(\hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}-\hat{j}+4 \hat{k}) \\\\ &=3 \hat{i}+3 \hat{j}-4 \hat{k} \end{aligned}$
$\begin{aligned} &D=\frac{1}{2}(\overrightarrow{A B} \times \overrightarrow{B C}) \end{aligned} \mid$
$=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{array}\right|$
$\begin{aligned} &=\hat{i}(12-3)+\hat{j}(-4-3)+\hat{k}(3+9) \\\\ &=9 \hat{i}+7 \hat{j}+12 \hat{k} \\\\ &=(\overrightarrow{A B} \times \overrightarrow{A C})=\sqrt{81+49+144} \\\\ &=\frac{1}{2} \sqrt{274} \text { square units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 36

Answer:
$\frac{2 \hat{i}-\hat{j}-\hat{k}}{\sqrt{6}},-\frac{3}{5} \hat{j}-\frac{4}{5} \hat{k}, \sqrt{404} \text { sq.units }$
Hint:
To solve this we use determinant method
Given:
$\begin{aligned} &\text { Magnitude }=10 \sqrt{3} \\ &\vec{a}=\hat{i}+2 \hat{j}+\hat{k} \\ &\vec{b}=-\hat{i}+3 \hat{j}+4 \hat{k} \end{aligned}$
Solution:
$\begin{aligned} &\vec{a}=\hat{i}+2 \hat{j}+\hat{k} \\\\ &\vec{b}=-\hat{i}+3 \hat{j}+4 \hat{k} \\\\ &\vec{a}+\vec{b}=\vec{c}=4 \hat{i}-2 \hat{j}-2 \hat{k} \end{aligned}$
$\begin{aligned} &|\vec{a} \times \vec{b}|=\frac{4 \hat{i}-2 \hat{j}-2 \hat{k}}{\sqrt{(4)^{2}+(-2)^{2}+(-2)^{2}}} \\\\ &=\frac{4 \hat{i}-2 \hat{j}-2 \hat{k}}{\sqrt{16+4+4}}=\frac{4 \hat{i}-2 \hat{j}-2 \hat{k}}{\sqrt{24}} \end{aligned}$
$\begin{aligned} &=\frac{4 \hat{i}-2 \hat{i}-2 \hat{k}}{2 \sqrt{6}} \\\\ &=\frac{2 \hat{i}-j-\hat{k}}{\sqrt{6}} \end{aligned}$
Similarly
$\begin{aligned} &\vec{d}=\vec{a}-\vec{b}=0 \hat{i}-6 \hat{j}-8 \hat{k} \\\\ &|\vec{a}-\vec{b}|=\frac{-6 \hat{j}-8 \hat{k}}{\sqrt{(-6)^{2}+(-8)^{2}}} \\\\ &=\frac{-6 \hat{j}-8 \hat{k}}{\sqrt{36+64}} \end{aligned}$
$\begin{aligned} &=\frac{-6 \hat{j}-8 \hat{k}}{10} \\\\ &=\frac{-3}{5} \hat{j}-\frac{4}{5} \hat{k} \end{aligned}$
Area of the parallelogram $=|\vec{a} \times \vec{b}|$
$=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & -5 \\ 2 & 2 & 3 \end{array}\right|$
$\begin{aligned} &=|-2 \hat{i}-16 \hat{j}+12 \hat{k}| \\\\ &=\sqrt{(-2)^{2}+(-16)^{2}+(12)^{2}} \\\\ &=\sqrt{4+256+144} \\\\ &=\sqrt{404} \text { sq.units } \end{aligned}$

Vector or Cross Product exercise 24.1 question 37

Answer:
4
Hint: To solve this we use $|\vec{a} \times \vec{b}|,|\vec{a} \cdot \vec{b}|$ formula
Given: $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=400$
Solution:
$\begin{aligned} &|\vec{a} \times \vec{b}|=|\vec{a}| \times|\vec{b}| \sin \theta \\\\ &|\vec{a} \cdot \vec{b}|=|\vec{a}| \times|\vec{b}| \cos \theta \\\\ &|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta=400 \end{aligned}$
$\begin{aligned} &|\vec{a}|^{2}|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=400 \\\\ &|\vec{a}|^{2}|\vec{b}|^{2}=400 \\\\ &|\vec{b}|^{2}=\frac{400}{|\vec{a}|^{2}}=\frac{400}{5^{2}}=\frac{400}{25} \end{aligned}$
$\begin{aligned} &|\vec{b}|^{2}=16 \\\\ &|\vec{b}|=4 \end{aligned}$

Vector or Cross Product exercise 24.1 question 38

Answer: $\frac{\sqrt{24}}{7}$
Hint: To solve this we use $\sin ^{2} \theta+\cos ^{2} \theta=1$ formula
Given:
$\begin{aligned} &\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \\\\ &\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k} \end{aligned}$
Solution:
$\begin{aligned} &\cos \theta=\vec{u} \cdot \vec{v}=|\vec{u}||\vec{v}| \cos \theta \\\\ &\cos \theta=\frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{\hat{i}-2 \hat{j}+3 \hat{k} \cdot(3 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{1^{2}+(-2)^{2}+(3)^{2}} \sqrt{3^{2}+(-2)^{2}+(1)^{2}}} \\\\ &\cos \theta=\frac{3+4+3}{\sqrt{14} \sqrt{14}}=\frac{10}{14} \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{5}{7} \\\\ &\sin ^{2} \theta+\cos ^{2} \theta=1 \\\\ &\sin ^{2} \theta=1-\cos ^{2} \theta \\\\ &\sin ^{2} \theta=1-\left(\frac{5}{7}\right)^{2} \end{aligned}$
$\begin{aligned} &\sin ^{2} \theta=\frac{49-25}{49} \\\\ &\sin ^{2} \theta=\frac{24}{49} \\\\ &\sin \theta=\sqrt{\frac{24}{49}}=\frac{\sqrt{24}}{7} \end{aligned}$

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