JEE Main 2023 — Mathematics Vectors & 3D Geometry

2023mcqeasy

Let $\vec{\alpha }=4\hat{i}+3\hat{j}+5\hat{k}$ and $\vec{\beta }=\hat{i}+2\hat{j}-4\hat{k}$ . Let ${\vec{\beta }}_{1}$ be parallel to $\vec{\alpha }$ and ${\vec{\beta }}_{2}$ be perpendicular to $\vec{\alpha }$ . If $\vec{\beta }={\vec{\beta }}_{1}+{\vec{\beta }}_{2}$ , then the value of $5{\vec{\beta }}_{2}\cdot (\hat{i}+\hat{j}+\hat{k})$ is

Official previous-year question

Held on 24 Jan 2023 · Verified 6 Jul 2026.

Options

A
$6$
B
$11$
C
$7$
D
$9$

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Solution

Given that $\vec{\alpha }=4\hat{i}+3\hat{j}+5\hat{k}$ and $\vec{\beta }=\hat{i}+2\hat{j}-4\hat{k}$ and $\vec{\beta }=\vec{{\beta }_{1}}+\vec{{\beta }_{2}}$

$\Rightarrow \vec{{\beta }_{2}}=\vec{\beta }-\vec{{\beta }_{1}}..........(1)$

since ${\vec{\beta }}_{1}$ is parallel to $\vec{\alpha }$ , $\vec{{\beta }_{1}}=t\vec{\alpha }$

$\Rightarrow \vec{{\beta }_{1}}=t(4\hat{i}+3\hat{j}+5\hat{k})=4t\hat{i}+3t\hat{j}+5t\hat{k}.........(2)$

Substituting the values of $\vec{{\beta }_{1}}$ and $\vec{\alpha }$ in $(1)$ , we get

$\vec{{\beta }_{2}}=\hat{i}+2\hat{j}-4\hat{k}-(4t\hat{i}+3t\hat{j}+5t\hat{k})=(1-4t)\hat{i}+(2-3t)\hat{j}+(-4-5t)\hat{k}............(3)$

since $\vec{{\beta }_{2}}$ is perpendicular to $\vec{\alpha }$ , so $\vec{{\beta }_{2}}\cdot \vec{\alpha }=0$

$\Rightarrow [(1-4t)\hat{i}+(2-3t)\hat{j}+(-4-5t)\hat{k}]\cdot [4\hat{i}+3\hat{j}+5\hat{k}]=0$

$\Rightarrow 4(1-4t)+3(2-3t)+5(-4-5t)=0$

$\Rightarrow 4-16t+6-9t-20-25t=0$

$\Rightarrow -50t=10$ $\Rightarrow t=\frac{-1}{5}$

From $(2)$ and $(3)$ , we get $\vec{{\beta }_{1}}=\frac{-1}{5}(4\hat{i}+3\hat{j}+5\hat{k})$

$\vec{{\beta }_{2}}=\frac{9}{5}\hat{i}+\frac{13}{5}\hat{j}-3\hat{k}=\frac{1}{5}(9\hat{i}+13\hat{j}-15\hat{k})$

So, the value of $5{\vec{\beta }}_{2}\cdot (\hat{i}+\hat{j}+\hat{k})$ will be,

$=5\times \frac{1}{5}(9\hat{i}+13\hat{j}-15\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=9+13-15=7$

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