JEE Main 2022 — Mathematics Vectors & 3D Geometry

2022mcqeasy

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that the angle between them is $\frac{\pi }{4}$ . If $\theta$ is the angle between the vectors $(\hat{a}+\hat{b})$ and $(\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}))$ then the value of $164{\mathrm{cos}}^{2}\theta$ is equal to

Official previous-year question

Held on 29 Jul 2022 · Verified 6 Jul 2026.

Options

A
$90+27\sqrt{2}$
B
$45+18\sqrt{2}$
C
$90+3\sqrt{2}$
D
$54+90\sqrt{2}$

Did you get this right?

Solution

Now let angle between $\hat{a}&\hat{b}=\frac{\pi }{4}=\phi$

So, $\hat{a}\cdot \hat{b}=|\hat{a}||\hat{b}|\mathrm{cos}\phi$

$\Rightarrow \hat{a}\cdot \hat{b}=\mathrm{cos}\phi =\frac{1}{\sqrt{2}}$

Now finding angle between $(\hat{a}+\hat{b})&(\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}))$ we get,

$\mathrm{cos}\theta =\frac{(\hat{a}+\hat{b})\cdot (\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b}))}{|\hat{a}+\hat{b}||\hat{a}+2\hat{b}+2(\hat{a}\times \hat{b})|}....(1)$

Now finding the value of ${|\hat{a}+\hat{b}|}^{2}=(\hat{a}+\hat{b})\cdot (\hat{a}+\hat{b})$

$\Rightarrow {|\hat{a}+\hat{b}|}^{2}=2+2\hat{a}\cdot \hat{b}$ $=2+\sqrt{2}$

Also value of $\hat{a}\times \hat{b}=|\hat{a}||\hat{b}|\mathrm{sin}\phi \hat{n}$

$\Rightarrow \hat{a}\times \hat{b}=\frac{\hat{n}}{\sqrt{2}}$ when $\hat{n}$ is vector $\perp$ to $\hat{a}$ and $\hat{b}$

Let $\vec{c}=\hat{a}\times \hat{b}$

We know that, $\vec{c}\cdot \vec{a}=0$ and $\vec{c}\cdot \vec{b}=0$ as perpendicular vector dot product is zero.

Now value of ${|\hat{a}+2\hat{b}+2\vec{c}|}^{2}$

$=1+4+\frac{(4)}{2}+4\hat{a}\cdot \hat{b}+8\hat{b}\cdot \vec{c}+4\vec{c}\cdot \hat{a}$

$=7+\frac{4}{\sqrt{2}}=7+2\sqrt{2}$

Now finding dot product we get, $(\hat{a}+\hat{b})\cdot (\hat{a}+2\hat{b}+2\vec{c})$

$={|\hat{a}|}^{2}+2\hat{a}\cdot \hat{b}+0+\hat{b}\cdot \hat{a}+2{|\hat{b}|}^{2}+0$

$=1+\frac{2}{\sqrt{2}}+\frac{1}{\sqrt{2}}+2$

$=3+\frac{3}{\sqrt{2}}$

Now putting all the values in equation $(1)$ we get,

$\mathrm{cos}\theta =\frac{3+\frac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}}\sqrt{7+2\sqrt{2}}}$

$\Rightarrow {\mathrm{cos}}^{2}\theta =\frac{9{(\sqrt{2}+1)}^{2}}{2(2+\sqrt{2})(7+2\sqrt{2})}$

$\Rightarrow {\mathrm{cos}}^{2}\theta =(\frac{9}{2\sqrt{2}})\frac{(\sqrt{2}+1)}{(7+2\sqrt{2})}$

$\Rightarrow 164{\mathrm{cos}}^{2}\theta =\frac{(82)(9)}{\sqrt{2}}\frac{(\sqrt{2}+1)}{(7+2\sqrt{2})}\frac{(7-2\sqrt{2})}{(7-2\sqrt{2})}$

$\Rightarrow 164{\mathrm{cos}}^{2}\theta =\frac{(82)}{\sqrt{2}}\frac{(9)[7\sqrt{2}-4+7-2\sqrt{2}]}{(41)}$

$\Rightarrow 164{\mathrm{cos}}^{2}\theta =(9\sqrt{2})[5\sqrt{2}+3]$

$\Rightarrow 164{\mathrm{cos}}^{2}\theta =90+27\sqrt{2}$

Your note

JEE Main Vectors & 3D Geometry in other years

Vectors & 3D Geometry 2026 Vectors & 3D Geometry 2025 Vectors & 3D Geometry 2024 Vectors & 3D Geometry 2023 Vectors & 3D Geometry 2021 Vectors & 3D Geometry 2020 Vectors & 3D Geometry 2019 Vectors & 3D Geometry 2018 Vectors & 3D Geometry 2017 Vectors & 3D Geometry 2016 Vectors & 3D Geometry 2015 Vectors & 3D Geometry 2014

More JEE Main Vectors & 3D Geometry questions

The shortest distance between the lines $\vec{r}=\left(\dfrac{1}{3}\hat{i}+2\hat{j}+\dfrac{8}{3}\hat{k}\right)+\lambda(2\hat{i}-5\hat{j}+6\hat{k})$ and $\vec{r}=\left(-\dfrac{2}{3}\hat{i}-\dfrac{1}{3}\hat{k}\right)+\mu(\hat{j}-\hat{k})$, $\lambda,\mu \in \mathbb{R}$, is:

2026medium

If the distance of the point $\mathrm{P}(43, \alpha, \beta), \beta<0$, from the line $\overrightarrow{\mathrm{r}}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k}), \mu \in \mathbf{R}$ along a line with direction ratios $3,-1,0$ is $13 \sqrt{10}$, then $\alpha^{2}+\beta^{2}$ is equal to $\_\_\_\_$

2026hard

The volume of the parallelepiped formed by vectors a=i+2j-k, b=2i-j+3k, c=3i+j+2k is:

2026hard

If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

2026hard

For a triangle ABC, let $\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{CA}}$ and $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{BA}}$. If $|\overrightarrow{\mathrm{p}}|=2 \sqrt{3},|\overrightarrow{\mathrm{q}}|=2$ and $\cos \theta=\frac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p} \times(\vec{q}-3 \vec{r})|^{2}+3|\vec{r}|^{2}$ is equal to :

2026hard

View all Vectors & 3D Geometry questions

Keep practicing Vectors & 3D Geometry

Work through every JEE Main Vectors & 3D Geometry PYQ, year by year.

Browse all Mathematics topics