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

Edited By Satyajeet Kumar | Updated on Jan 27, 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.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

This Story also Contains

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:
$\begin{aligned} &\cos \theta=\vec{u} \cdot \vec{v}=|\vec{u}||\vec{v}| \cos \theta \\\\ &\cos \theta=\frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{\hat{i}-2 \hat{j}+3 \hat{k} \cdot(3 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{1^{2}+(-2)^{2}+(3)^{2}} \sqrt{3^{2}+(-2)^{2}+(1)^{2}}} \\\\ &\cos \theta=\frac{3+4+3}{\sqrt{14} \sqrt{14}}=\frac{10}{14} \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{5}{7} \\\\ &\sin ^{2} \theta+\cos ^{2} \theta=1 \\\\ &\sin ^{2} \theta=1-\cos ^{2} \theta \\\\ &\sin ^{2} \theta=1-\left(\frac{5}{7}\right)^{2} \end{aligned}$
$\begin{aligned} &\sin ^{2} \theta=\frac{49-25}{49} \\\\ &\sin ^{2} \theta=\frac{24}{49} \\\\ &\sin \theta=\sqrt{\frac{24}{49}}=\frac{\sqrt{24}}{7} \end{aligned}$

Experts prepare Class 12th RD Sharma Chapter 24 Exercise 24.1 Solutions using the latest CBSE guidelines and marking schemes. When a student refers to RD Sharma Class 12th Solutions Vector or Cross Product Ex. 24.1, it helps them have a clear understanding of the topic. When a student relates to RD Sharma Class 12th Exercise 24.1, the following benefits will accrue:

Expertly crafted:

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

A team of subject experts creates RD Sharma Class 12 Solutions Chapter 24 ex 24.1 in accordance with the most recent marking pattern and CBSE guidelines. Every student must practise RD Sharma Class 12th Exercise 24.1 in order to achieve a high score. RD Sharma Class 12th Chapter 24 Exercise 24.1 solutions explain every topic so that every student can learn and solve it.

2. The best source of preparation:

Because of their high quality, RD Sharma Class 12th Exercise 24.1 solutions are the best to use. The most recent NCERT and CBSE updates make it an excellent choice. Furthermore,if the practice is done regularly, it would be easier to score good marks.

3. Questions from the NCERT:

The NCERT is usually used to set questions in exams. Teachers assign question papers based on NCERT textbooks, so it is necessary to learn NCERT concepts and solve questions from RD Sharma Class 12 Solutions.

4. Various approaches to answering a question

When a team develops a solution,multiple approaches towards a single question develops making it easier for a student to understand the concept. The experts have devised novel approaches to answering a single question, making us unique.

5. Performance as a benchmark:

RD Sharma Class 12 Solutions Chapter 24 ex 24.1 solutions help students perform better on exams. RD Sharma Class 12 Solutions Ex. 24.1 covers all of the essential questions found in exams, allowing students to establish a benchmark score.

6. No cost:

Students will find all of these solutions for free at Career360, the best website to visit for all of the benefits associated with board exam questions and competitive exams. With Career360, a student will understand, learn, and perform exceptionally well in exams. These solutions have benefited thousands of students, making Career360 the ideal choice for all students. If you still have a doubt, visit the website and access it once. You will surely love it.

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. What is the formula cross product?

If two vectors A and B have an angle θ, then the formula for the cross product of vectors is given by:

A × B = |A| |B| sin θ

2. What do you mean by a cross product?

A Cross product is a binary operation on two vectors in three-dimensional space.

3. How can I avail the benefit from these solutions?

You can visit the website, download the PDF for free and practice regularly to score good marks and avail full benefits.

4. Will the RD Sharma Solutions for Class 12 help to score brilliant marks in the board exams?

Yes, the RD Sharma Solutions for Class 12 is one of the best study materials which are available online. Chapter-wise and Exercise-wise solutions can be referred to by the students when they are not able to find or understand an accurate answer for the textbook questions.

5. How can I get an understanding of the key concepts in the RD Sharma Class 12th Exercise 24.1?

Before the start of an academic year, every student should download and read the CBSE syllabus to get a good understanding of the concepts that will be covered in the exams.