Vector Algebra · JEE Main previous year questions

2024 31 questions

1. Q22024 · 29 Jan 2024 · Shift 1

   Let $\vec{a},\vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear. If $\vec{a}+5\vec{b}$ is collinear with $\vec{c}$, $\vec{b}+6\vec{c}$ is collinear with $\vec{a}$ and $\vec{a}+\alpha \vec{b}+\beta \vec{c}=\vec{0}$, then $\alpha +\beta$ is equal to

   1. A.$30$
   2. B.$-30$
   3. C.$-25$
   4. D.$35$
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   Answer: D

2. Q112024 · 29 Jan 2024 · Shift 1

   Let $O$ be the origin and the position vectors of $A$ and $B$ be $2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}+4\hat{k}$ respectively. If the internal bisector of $\angle \text{AOB}$ meets the line AB at C, then the length of $OC$ is

   1. A.$\frac{3}{2}\sqrt{34}$
   2. B.$\frac{2}{3}\sqrt{31}$
   3. C.$\frac{2}{3}\sqrt{34}$
   4. D.$\frac{3}{2}\sqrt{31}$
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   Answer: C

3. Q52024 · 29 Jan 2024 · Shift 2

   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 $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k},\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$ and $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$ 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

   1. A.$\frac{11}{2}$
   2. B.$\frac{5}{2}$
   3. C.7
   4. D.9
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   Answer: B

4. Q182024 · 27 Jan 2024 · Shift 1

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

   1. A.32
   2. B.36
   3. C.24
   4. D.20
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   Answer: C

5. Q22024 · 27 Jan 2024 · Shift 2

   Let the position vectors of the vertices A, B and C of a triangle be $2\hat{i}+2\hat{j}+\hat{k}$, $\hat{i}+2\hat{j}+2\hat{k}$ and $2\hat{i}+\hat{j}+2\hat{k}$ respectively. Let $l_1,l_2$ and $l_3$ be the lengths of perpendiculars drawn from the orthocentre of the triangle on the sides AB, BC and CA respectively, then $l_1^2+l_2^2+l_3^2$ equals:

   1. A.$\frac{1}{4}$
   2. B.$\frac{1}{5}$
   3. C.$\frac{1}{3}$
   4. D.$\frac{1}{2}$
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   Answer: D

6. Q202024 · 27 Jan 2024 · Shift 2

   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 $l$ denote the length of the angle bisector AD of $\angle \text{BAC}$ where D is on the line segment BC, then $2l^2$ equals :

   1. A.$45$
   2. B.$50$
   3. C.$42$
   4. D.$49$
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   Answer: A