Vector Algebra Previous Year Questions with Solutions

1.0Introduction

Vector Algebra Previous Year Questions typically cover topics like vector addition, scalar multiplication, dot product, cross product, and applications of vectors in geometry and physics. Examples include finding the angle between two vectors, determining the magnitude of a vector, proving vector identities, and solving problems related to displacement and velocity. Solutions involve applying vector formulas and properties such as vector projection, triple products, and the geometric interpretation of vectors. Practicing these questions enhances understanding of vector operations and improves problem-solving efficiency, helping students gain confidence and prepare effectively for exams.

2.0Vector Algebra Previous Year Questions for JEE with Solutions

JEE questions in Vector Algebra often test concepts related to vector operations, scalar and vector products, and geometric interpretations of vectors. Some common types of problems include:

• Basic Vector Operations: Questions on vector addition, subtraction, scalar multiplication, and finding unit vectors.
• Dot and Cross Products: Problems involving the angle between two vectors, projection of one vector on another, and checking perpendicularity or parallelism using dot or cross product.
• Applications in Geometry: Questions related to position vectors, section formula, finding the area of a triangle using cross product, and vector equations of lines.

These questions are aimed at testing conceptual clarity, vector manipulation skills, and geometric reasoning.

Note: In the JEE Main Mathematics exam, you can generally expect 2 to 3 questions from the Vector Algebra chapter.

3.0JEE Questions in Vector Algebra: Key Concepts

1. Basic Vector Operations

• Vector from point $A(x_1,y_1,z_1)$ to $B(x_2,y_2,z_2):$

$\overrightarrow{AB}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

• Magnitude of vector:

$|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$

• Direction cosines of a vector:

$\cos \alpha =\frac{a_1}{|\vec{a}|},\cos \beta =\frac{a_2}{|\vec{a}|},\cos \gamma =\frac{a_3}{|\vec{a}|}$

• Relation:

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

2. Scalar (Dot) Product

• If , then vectors are perpendicular.

$$\begin{gathered} \vec{a}\cdot \vec{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3 \\ \vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta \\ If\vec{a}\cdot \vec{b}=0,\text{then vectors are perpendicular}. \end{gathered}$$

3. Vector (Cross) Product

• If , vectors are parallel. $$\begin{gathered} \vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right| \\ |\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta \\ If\vec{a}\times \vec{b}=\vec{0},\text{ vectors are parallel}. \end{gathered}$$

4. Scalar Triple Product (STP)

• Volume of a parallelepiped
• If STP = 0 → vectors are coplanar
• $$\begin{gathered} \vec{a}\cdot (\vec{b}\times \vec{c})=\text{Volume of a parallelepiped} \\ IfSTP=0\Rightarrow \text{vectors are coplanar} \end{gathered}$$

5. Vector Triple Product Identity

$\vec{a}\times (\vec{b}\times \vec{c})=(\vec{a}\cdot \vec{c})\vec{b}-(\vec{a}\cdot \vec{b})\vec{c}$

6. Section Formula (Vectors)

• If P divides in ratio m:n,
• Midpoint:

$$\begin{gathered} \text{If P divides }\overrightarrow{AB}\text{in ratio m:n}, \\ \overrightarrow{OP}=\frac{m\vec{B}+n\vec{A}}{m+n} \\ \text{Midpoint}: \\ \vec{M}=\frac{\vec{A}+\vec{B}}{2} \end{gathered}$$

7. Collinearity of Vectors

• Vectors a,b are collinear if:

$\vec{a}=k\vec{b}$

or

$\vec{a}\times \vec{b}=0$

8. Vector Equation of a Line 

$$\begin{gathered} \text{Through point}\vec{a},parallelto\vec{b}: \\ \vec{r}=\vec{a}+\lambda \vec{b} \\ \text{Through two points}\vec{a},\vec{b}: \\ \vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a}) \end{gathered}$$

9. Angle Between Two Vectors

$$\begin{gathered} \cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|} \\ \sin \theta =\frac{|\vec{a}\times \vec{b}|}{|\vec{a}||\vec{b}|} \end{gathered}$$

10. Area & Volume

$$\begin{gathered} \text{Area of triangle formed by vectors }\vec{a},\vec{b}: \\ A=\frac{1}{2}|\vec{a}\times \vec{b}| \\ \text{Volume of parallelepiped formed by }\vec{a},\vec{b},\vec{c}: \\ V=|\vec{a}\cdot (\vec{b}\times \vec{c})| \end{gathered}$$

11. Coplanarity Condition

$$\begin{gathered} \text{Vectors}\vec{a},\vec{b},\vec{c}\text{are coplanar if:} \\ \vec{a}\cdot (\vec{b}\times \vec{c})=0 \end{gathered}$$

12. Shortcuts & Tricks

• If three points A, B, C are collinear:

$\overrightarrow{AB}\times \overrightarrow{AC}=0$

• If point lies on a line:
• Substitute in vector form and satisfy the equation.

4.0JEE Mains Past Year Questions with Solutions on Vector Algebra 

1. If λ > 0, let θ be the angle between the vectors $\vec{a}=\hat{i}+\lambda \hat{j}-3\hat{k}and\vec{b}=3\hat{i}-\hat{j}+2\hat{k}.$ If the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are mutually perpendicular, then the value of (14 cos θ)² is equal to 

(1) 25

(2) 20

(3) 50

(4) 40

Ans. (1)

Solution:

$$\begin{gathered} (\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=0,\lambda >0 \\ |\vec{a}{|}^2-|\vec{b}{|}^2=0\Rightarrow 1+{\lambda }^2+9=9+1+4 \\ \cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{3-\lambda -6}{\sqrt{14}\cdot \sqrt{14}} \\ \Rightarrow \lambda =2 \\ 14\cos \theta =3-8=-5 \\ \Rightarrow (14\cos \theta {)}^2=25 \end{gathered}$$

2. Let a unit vector which makes an angle of 60° with $\overrightarrow{2i}+\overrightarrow{2j}-\vec{k}\text{and an angle of 45° with}\vec{i}-\vec{k}be\vec{C}$. Then $\vec{C}=\left(\frac{\sqrt{2}}{3}\right)\vec{i}-\left(\frac{1}{2}\right)\vec{k}is:$

$$\begin{gathered} (1)\frac{\sqrt{2}}{3}\vec{i}+\frac{1}{3\sqrt{2}}\vec{j}-\frac{1}{2}\vec{k} \\ (2)\left(\frac{1}{2}\vec{i}+\frac{1}{3\sqrt{2}}\vec{j}+\left(\frac{2\sqrt{2}}{3}\right)\vec{k}\right) \\ (3)\left(\frac{1}{\sqrt{3}}+\frac{1}{2}\right)\vec{i}+\left(\frac{1}{\sqrt{3}}-\frac{1}{3\sqrt{2}}\right)\vec{j}+\left(\frac{1}{\sqrt{3}}\cdot \frac{\sqrt{2}}{3}\right)\vec{k} \\ (4)\frac{\sqrt{2}}{3}\vec{i}-\frac{1}{2}\vec{k} \end{gathered}$$

Ans. (4)

Solution: 

$$\begin{gathered} \vec{C}=C_1\vec{i}+C_2\vec{j}+C_3\vec{k} \\ {C}_1^2+C_2^2+C_3^2=1 \\ \vec{C}\cdot (2\vec{i}+2\vec{j}-\vec{k})=|\vec{C}||\vec{v}|\cos 60^{\circ } \\ 2C_1+2C_2-C_3=\frac{3}{2} \\ {C}_1-C_3=1 \\ {C}_1+2C_2=\frac{1}{2} \\ {C}_1=\frac{\sqrt{2}}{3}+\frac{1}{2},C_2=-\frac{1}{3\sqrt{2}},C_3=\frac{\sqrt{2}}{3}-\frac{1}{2} \\ \overrightarrow{AB}=\vec{i}+2\vec{j}-7\vec{k},\overrightarrow{BC}=a\vec{i}+b\vec{j}+c\vec{k} \\ and \\ \overrightarrow{AC}=6\vec{i}+d\vec{j}-2\vec{k},d>0 \end{gathered}$$

3. Let ABC be a triangle of area $15\sqrt{2}$ and the vectors $\overrightarrow{AB}=\hat{i}+2\hat{j}-7\hat{k},\overrightarrow{BC}=a\hat{i}+b\hat{j}+c\hat{k},\overrightarrow{AC}=6\hat{i}+d\hat{j}-2\hat{k},d>0.,d>0$. Then the square of the length of the largest side of the triangle ABC is

Ans. (54)

Solution: 

$$\begin{gathered} \text{Area}=\frac{1}{2}\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 12-7\\ 6d-2\end{array}\right|=15\sqrt{2} \\ =\frac{1}{2}\left[(-4+7d)\hat{i}-\hat{j}(-2+42)+\hat{k}(d-12)\right] \\ =\frac{1}{2}\left[(7d-4)\hat{i}+40\hat{j}+(d-12)\hat{k}\right] \\ \Rightarrow \text{Magnitude}=\frac{1}{2}\sqrt{(7d-4{)}^2+40^2+(d-12{)}^2}=15\sqrt{2} \\ \Rightarrow (7d-4{)}^2+1600+(d-12{)}^2=1800 \\ \Rightarrow 49d^2-56d+16+1600+d^2-24d+144=1800 \\ \Rightarrow 50d^2-80d+1760=1800 \\ \Rightarrow 50d^2-80d-40=0 \\ \Rightarrow 5d^2-8d-4=0 \\ \Rightarrow 5d^2-10d-2d-4=0 \\ \Rightarrow 5d(d-2)+2(d-2)=0 \\ \Rightarrow (5d+2)(d-2)=0 \\ \Rightarrow d=2\text{or}d=-\frac{2}{5} \\ \Rightarrow d>0\Rightarrow d=2\text{Now solve for vector}\overrightarrow{BC}:\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=(6-1)\hat{i}+(2-2)\hat{j}+(-2+7)\hat{k}=5\hat{i}+0\hat{j}+5\hat{k} \\ \text{So comparing}\overrightarrow{BC}=a\hat{i}+b\hat{j}+c\hat{k}: \\ a+1=6\Rightarrow a=5,b+2=2\Rightarrow b=0,c-7=-2\Rightarrow c=5 \\ \text{Length of sides}|\overrightarrow{AB}|=\sqrt{1^2+2^2+(-7{)}^2}=\sqrt{1+4+49}=\sqrt{54} \\ |\overrightarrow{BC}|=\sqrt{5^2+0^2+5^2}=\sqrt{25+25}=\sqrt{50} \\ |\overrightarrow{AC}|=\sqrt{6^2+2^2+(-2{)}^2}=\sqrt{36+4+4}=\sqrt{44} \end{gathered}$$

4. Let $\vec{a}=\hat{i}-3\hat{j}+7\hat{k},\vec{b}=2\hat{i}-\hat{j}+\hat{k}and\vec{c}$ be a vector such that $(\vec{a}+2\vec{b})\times \vec{c}=3(\vec{c}\times \vec{a})$. If $\vec{a}\cdot \vec{c}=130,$ then $\vec{b}\cdot \vec{c}$ is equal to __________.

Ans. (30)

Solution: 

$$\begin{gathered} (\vec{a}+2\vec{b})\times \vec{c}=3(\vec{c}\times \vec{a}) \\ (2\vec{b}+4\vec{a})\times \vec{c}=0 \\ \vec{c}=\lambda (4\vec{a}+2\vec{b})=\lambda (8\hat{i}-14\hat{j}+30\hat{k}) \\ \vec{a}\cdot \vec{c}=130 \\ 8\lambda +42\lambda +210\lambda =130 \\ \lambda =\frac{1}{2} \\ \vec{c}=4\hat{i}-7\hat{j}+15\hat{k} \\ \vec{b}\cdot \vec{c}=8+7+15=30 \end{gathered}$$

5. Let $\vec{a}=2\hat{i}+\hat{j}-\hat{k},\vec{b}=\left((\vec{a}\times (\hat{i}+\hat{j}))\times \hat{i}\right)\times \hat{i}.$ Then the square of the projection of $\vec{a}on\vec{b}$ is :

$$\begin{gathered} (1)\frac{1}{5} \\ (2)2 \\ (3)\frac{1}{3} \\ (4)\frac{2}{3} \end{gathered}$$

Ans. (2)

Solution: 

$$\begin{gathered} \vec{a}\times (\hat{i}+\hat{j})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 1& 1& 0\end{array}\right| \\ =\hat{i}-\hat{j}+\hat{k} \\ ((\vec{a}\times (\hat{i}+\hat{j}))\times \hat{i})=\hat{k}+\hat{j} \\ (((\vec{a}\times (\hat{i}+\hat{j}))\times \hat{i})\times \hat{i})=\hat{j}-\hat{k} \\ \text{projection of }\vec{a}\text{ on }\vec{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|} \\ =\frac{1+1}{\sqrt{2}}=\sqrt{2} \end{gathered}$$

6. Let $\vec{a}=4\hat{i}-\hat{j}+\hat{k},\vec{b}=11\hat{i}-\hat{j}+\hat{k},\vec{c}$ be a vector such that $(\vec{a}+\vec{b})\times \vec{c}=\vec{c}\times (-2\vec{a}+3\vec{b})$. If $(2\vec{a}+3\vec{b})\cdot \vec{c}=1670,$ then $|\vec{c}{|}^2$ is equal to:

(1) 1627

(2) 1618

(3) 1600

(4) 1609

Ans. (2)

Solution: 

$$\begin{gathered} (\vec{a}+\vec{b})\times \vec{c}=\vec{c}\times (-2\vec{a}+3\vec{b})=0 \\ (\vec{a}+\vec{b})\times \vec{c}+(2\vec{a}-3\vec{b})\times \vec{c}=0 \\ \Rightarrow (\vec{a}+\vec{b})-2\vec{a}+3\vec{b})\times \vec{c}=0 \\ \Rightarrow \vec{c}=\lambda (4\vec{b}-\vec{a}) \\ =\lambda (44\hat{i}-4\hat{j}+4\hat{k}-4\hat{i}+\hat{j}-\hat{k}) \\ =\lambda (40\hat{i}-3\hat{j}+3\hat{k}) \\ (8\hat{i}-2\hat{j}+2\hat{k}+33\hat{i}-3\hat{j}+3\hat{k})\cdot (40\hat{i}-3\hat{j}+3\hat{k})=1670 \\ (4\hat{i}-5\hat{j}+5\hat{k})\cdot (40\hat{i}-3\hat{j}+3\hat{k})\lambda =1670 \\ \Rightarrow (1640+15+15)\lambda =1670\Rightarrow \lambda =1 \\ \Rightarrow \vec{c}=40\hat{i}-3\hat{j}+3\hat{k}\Rightarrow |\vec{c}{|}^2=1600+9+9=\boxed{1618} \end{gathered}$$

7. Let $\vec{a}=2\hat{i}+\alpha \hat{j}+\hat{k},\vec{b}=-\hat{i}+\hat{k},\vec{c}=\beta \hat{j}-\hat{k}$ where α and β are integers and αβ = –6. Let the values of the ordered pair (α, β) for which the area of the parallelogram of diagonals $\vec{a}+\vec{b}and\vec{b}+\vec{c}is\sqrt{21}2$, be (α₁, β₁) and (α₂, β₂). Then ${\alpha }_1^2+{\beta }_1^2$ – α₂β₂ is equal to 

(1) 17

(2) 24

(3) 21

(4) 19

Ans. (4)

Solution: 

$$\begin{gathered} \text{Area of parallelogram}=\frac{1}{2}|\overrightarrow{d_1}\times \overrightarrow{d_2}| \\ A=\frac{1}{2}|(\vec{a}+\vec{b})\times (\vec{b}+\vec{c})|=\frac{\sqrt{21}}{2} \\ so,\vec{a}+\vec{b}=\hat{i}+\alpha \hat{j}+2\hat{k} \\ \vec{b}+\vec{c}=-\hat{i}+\beta \hat{j} \\ (\vec{a}+\vec{b})\times (\vec{b}+\vec{c})=\left|\begin{array}{cc}\hat{i}& \hat{j}\hat{k}\\ 1\alpha 2& \\ -1\beta 0& \end{array}\right| \\ =\hat{i}(-2\beta )-\hat{j}(2)+\hat{k}(\beta +\alpha ) \\ =-2\beta \hat{i}-2\hat{j}+(\beta +\alpha )\hat{k} \\ |(\vec{a}+\vec{b})\times (\vec{b}+\vec{c})|=\sqrt{4{\beta }^2+4+(\beta +\alpha {)}^2}=\sqrt{21} \\ \Rightarrow 4{\beta }^2+4+{\alpha }^2+{\beta }^2+2\alpha \beta =21 \\ \Rightarrow {\alpha }^2+5{\beta }^2+2\alpha \beta =17 \\ and\alpha \beta =-6 \\ \alpha =-3,\beta =2\Rightarrow ({\alpha }_1,{\beta }_1)=(-3,2) \\ \alpha =3,\beta =-2\Rightarrow ({\alpha }_2,{\beta }_2)=(3,-2) \\ {\alpha }_1^2+{\beta }_1^2-{\alpha }_2{\beta }_2=9+4+6=\boxed{19} \end{gathered}$$

8. The position vectors of the vertices A, B and C of a triangle are $2\hat{i}-3\hat{j}+3\hat{k},2\hat{i}+2\hat{j}+3\hat{k},and-\hat{i}+\hat{j}+3\hat{k}$ respectively. Let I denotes the length of the angle bisector AD of $\angle BAC$ where D is on the line segment BC, then 2l² equals: 

1) 49

(2) 42

(3) 50

(4) 45

Ans. (4)

Solution: 

AB = 5

AC = 5

D is midpoint of BC

$$\begin{gathered} \vec{D}=\left(\frac{1}{2},\frac{3}{2},3\right) \\ l=\sqrt{(2-\frac{1}{2}{)}^2+(-3-\frac{3}{2}{)}^2+(3-3{)}^2} \\ =\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(-\frac{9}{2}\right)}^2} \\ =\sqrt{\frac{9}{4}+\frac{81}{4}}=\sqrt{\frac{90}{4}}=\sqrt{\frac{45}{2}} \\ 2l^2=\boxed{45} \end{gathered}$$

9. Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}，\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$. Let $\vec{c}$ be the vector such that $\vec{a}\times \vec{c}=\vec{b}$ and $\vec{a}\cdot \vec{c}=3.$

 Solution: 

$$\begin{gathered} \vec{a}\cdot \left[(\vec{c}\times \vec{b})-\vec{b}-\vec{c}\right]=\vec{a}\cdot (\vec{c}\times \vec{b})-\vec{a}\cdot \vec{b}-\vec{a}\cdot \vec{c}\dots ..(i)given,\vec{a}\times \vec{c}=\vec{b} \\ (\vec{a}\times \vec{c})\cdot \vec{b}=|\vec{b}{|}^2=27\Rightarrow \vec{a}\cdot (\vec{c}\times \vec{b})=27\dots \dots (ii) \\ \vec{a}\cdot \vec{b}=3-6+3=0\dots ..(iii) \\ \vec{a}\cdot \vec{c}=3\dots ..(iv) \\ \vec{a}\cdot \left[(\vec{c}\times \vec{b})-\vec{b}-\vec{c}\right]=27-0-3=\boxed{24} \end{gathered}$$

10. Let a unit vector $\hat{u}=x\hat{i}+y\hat{j}+z\hat{k}$ make angles $\frac{\pi }{2},\frac{\pi }{3}and\frac{2\pi }{3}$with the vectors $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k},\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j},\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$ respectively. If $\vec{v}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k},$

then $|\hat{u}-\vec{v}{|}^2$ is equal to

$$\begin{gathered} (1)\frac{11}{2} \\ (2)\frac{5}{2} \\ (3)9 \\ (4)7 \end{gathered}$$Ans. (2)

Solution: 

$$\begin{gathered} \text{Unit vector }\hat{u}=x\hat{i}+y\hat{j}+z\hat{k} \\ {\vec{\beta }}_1=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}, \\ {\vec{\beta }}_2=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}, \\ {\vec{\beta }}_3=\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k} \\ \text{Now angle between }\hat{u}{\vec{\beta }}_1=\frac{\pi }{2} \\ \hat{u}\cdot {\vec{\beta }}_1=0\Rightarrow \frac{x}{\sqrt{2}}+\frac{z}{\sqrt{2}}=0\Rightarrow x+z=0\dots \dots (i) \\ \hat{u}{\vec{\beta }}_2\frac{\pi }{3}： \\ \hat{u}\cdot {\vec{\beta }}_2=|\hat{u}||{\vec{\beta }}_2|\cos \frac{\pi }{3} \\ \Rightarrow \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} \\ \Rightarrow y+z=\frac{1}{\sqrt{2}}\dots ..(ii) \\ \hat{u}{\vec{\beta }}_3\frac{2\pi }{3}： \\ \hat{u}\cdot {\vec{\beta }}_3=|\hat{u}||{\vec{\beta }}_3|\cos \frac{2\pi }{3} \\ \Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=-\frac{1}{2} \\ \Rightarrow x+y=-\frac{1}{\sqrt{2}}\dots \dots .(iii) \\ \text{from equation (i), (ii) and (iii) we get } \\ x=-\frac{1}{\sqrt{2}},y=0,z=\frac{1}{\sqrt{2}} \\ Thus\hat{u}-\vec{v}=-\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}-\left(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}\right)=-\frac{2}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j} \\ \therefore |\hat{u}-\vec{v}{|}^2={\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)}^2=\frac{5}{2} \end{gathered}$$

11. Let $\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k},\alpha ,\beta \in \mathbb{R}$. Let a vector $\vec{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi }{4}$ and $|\vec{b}|=6,$ If $\vec{a}\cdot \vec{b}=3\sqrt{2}$, then the value of $({\alpha }^2+{\beta }^2)|\vec{a}\times \vec{b}{|}^2$ is equal to 

(1) 90

(2) 75

(3) 95

(4) 85

Ans. (1)

Solution: 

$$\begin{gathered} |\vec{b}|=6,|\vec{a}|\cdot |\vec{b}|\cos \theta =3\sqrt{2} \\ \Rightarrow \vec{a}\cdot \vec{b}=18 \\ |\vec{a}{|}^2=6 \\ Also1+{\alpha }^2+{\beta }^2=6 \\ {\alpha }^2+{\beta }^2=5 \\ \text{ to find } \\ ({\alpha }^2+{\beta }^2)\cdot |\vec{a}{|}^2\cdot |\vec{b}{|}^2\cdot {\sin }^2\theta \\ =(5)(6)(6)\cdot \left(\frac{1}{2}\right) \\ =\boxed{90} \end{gathered}$$

12. Let $\vec{a}and\vec{b}$ be two vectors such that and $|\vec{a}|=1,|\vec{b}|=4,and\vec{a}\cdot \vec{b}=2.$ If $\vec{c}=(2\vec{a}\times \vec{b})-3\vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is α, then 192sin²α is equal to______

Ans. (48)

Solution: 

$$\begin{gathered} \vec{b}\cdot \vec{c}=(2\vec{a}\times \vec{b})\cdot \vec{b}-3|\vec{b}{|}^2 \\ |\vec{b}||\vec{c}|\cos \alpha =-3|\vec{b}{|}^2 \\ \Rightarrow |\vec{c}|\cos \alpha =-12,\text{as }|\vec{b}|=4 \\ \vec{a}\cdot \vec{b}=2 \\ \Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3} \\ |\vec{c}{|}^2=|(2\vec{a}\times \vec{b})-3\vec{b}{|}^2 \\ =4|\vec{a}\times \vec{b}{|}^2+9|\vec{b}{|}^2 \\ =64\cdot \frac{3}{4}+144=192 \\ |\vec{c}{|}^2{\cos }^2\alpha =144 \\ \Rightarrow 192{\cos }^2\alpha =144 \\ \Rightarrow 192{\sin }^2\alpha =\boxed{48} \end{gathered}$$