Consider two vectors u =3 i - j and v =2 i + j - k , >0. The angl… - Vidyadip

MCQ

Consider two vectors $\vec{\mathrm{u}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}$ and $\vec{\mathrm{v}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}, \lambda>0$. The angle between them is given by $\cos ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)$. Let $\vec{v}=\vec{v}_{1}+\vec{v}_{2}$, where $\vec{v}_{1}$ is parallel to $\vec{\mathrm{u}}$ and $\vec{\mathrm{v}}_{2}$ is perpendicular to $\vec{\mathrm{u}}$. Then the value $\left|\vec{v}_{1}\right|^{2}+\left|\vec{v}_{2}\right|^{2}$ is equal to

A
$\frac{23}{2}$
✓
$14$
C
$\frac{25}{2}$
D
$10$

✓

Answer

Correct option: B.

$14$

(B) 14
$\vec{\mathrm{u}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}, \vec{\mathrm{v}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\lambda \hat{\mathrm{k}}$,
$\Rightarrow \frac{\vec{\mathrm{u}} \cdot \vec{\mathrm{v}}}{|\vec{\mathrm{u}}||\vec{\mathrm{v}}|}=\cos \theta$
$\Rightarrow \frac{5}{\sqrt{10} \sqrt{5+\lambda^{2}}}=\frac{\sqrt{5}}{2 \sqrt{7}}$
$\Rightarrow \lambda^{2}=9 \quad \Rightarrow \lambda=3(\because \lambda>0)$
$\vec{\mathrm{v}}=\vec{\mathrm{v}}_{1}+\vec{\mathrm{v}}_{2}$
$\Rightarrow|\vec{\mathrm{v}}|^{2}=\vec{\mathrm{v}}_{1}^{2}+\vec{\mathrm{v}}_{2}^{2}+2 \vec{\mathrm{v}}_{1} \cdot \vec{\mathrm{v}}_{2}$
$\Rightarrow 14=\vec{\mathrm{v}}_{1}^{2}+\vec{\mathrm{v}}_{2}^{2}+0 \quad\left(\because \vec{\mathrm{v}}_{1} \perp \vec{\mathrm{v}}_{2}\right)$
$\Rightarrow\left|\vec{\mathrm{v}}_{1}^{2}\right|+\left|\vec{\mathrm{v}}_{1}^{2}\right|=14$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "SECTION - A [MATHS - MCQ]" from JEE Main 4-April-2025 Paper - Shift 1→JEE Main 4-April-2025 Paper - Shift 1 — all question types→JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The equation of the line joining the point $(3, 5)$ to the point of intersection of the lines $4x + y - 1 = 0$ and $7x - 3y - 35 = 0$ is equidistant from the points $(0, 0)$ and $(8, 34)$

If $\frac{{1 - {{\tan }^2}\theta }}{{{{\sec }^2}\theta }} = \frac{1}{2}$, then the general value of $\theta $ is

If $\sum_{ r =1}^9\left(\frac{ r +3}{2^r}\right) \cdot{ }^9 C _{ r }=\alpha\left(\frac{3}{2}\right)^9-\beta, \quad \alpha, \beta \in N , \quad$ then $(\alpha+\beta)^2$ is equal to

From a lot of $12$ items containing $3$ defectives, a sample of $5$ items is drawn at random. Let the random variable $\mathrm{X}$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ is equal to..........

If $\int_0^{\frac{\pi}{3}} \cos ^4 x d x=a \pi+b \sqrt{3}$, where $a$ and $b$ are rational numbers, then $9 a +8 b$ is equal to $:$

The probability of $A, B, C$ solving a problem are $\frac{1}{3},\,\frac{2}{7},\,\frac{3}{8}$ respectively. If all the three try to solve the problem simultaneously, the probability that exactly one of them will solve it, is

Middle point of the chord of the circle ${x^2} + {y^2} = 25$ intercepted on the line $x - 2y = 2$ is

Two tangents are drawn from the point $\mathrm{P}(-1,1)$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-6 \mathrm{y}+6=0$. If these tangents touch the circle at points $A$ and $B$, and if $D$ is a point on the circle such that length of the segments $A B$ and $A D$ are equal, then the area of the triangle $A B D$ is eqaul to:

Probability of throwing $16$ in one throw with three dice is

Let the area of the region $\left\{(x, y): 2 y \leq x^{2}+3\right.$, $y+|x| \leq 3, y \geq|x-1|\}$ be A. Then $6 A$ is equal to: