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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 4 · Set 1 · Q25
(a) Let two rods placed on the ground be represented by vectors 4 i - j +3 k and -2 i + j -2 k. Find a vector representing a flag-post of height 5 m…
(a)
Let two rods placed on the ground be represented by vectors $\displaystyle 4 \hat{i}-\hat{j}+3 \hat{k}$ and $\displaystyle -2 \hat{i}+\hat{j}-2 \hat{k}$. Find a vector representing a flag-post of height $\displaystyle 5$ m that has to be erected perpendicular to both the rods.
OR
(b)
A unit vector $\displaystyle \overrightarrow{\mathrm{a}}$ is such that it makes an angle $\displaystyle \frac{\pi}{4}$ with x-axis, $\displaystyle \frac{\pi}{3}$ with y-axis and an acute angle $\displaystyle \theta$ with z-axis. Find $\displaystyle \theta$ and the components of $\displaystyle \overrightarrow{\mathrm{a}}$.
Marking-scheme solution
Let $\displaystyle \vec{a}=4 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=-2 \hat{i}+\hat{j}-2 \hat{k}$
$\displaystyle \vec{a} \times \vec{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix}=-\hat{i}+2 \hat{j}+2 \hat{k}$
Now, Required vector $\displaystyle =5\left(\dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\right)=\dfrac{-5}{3} \hat{i}+\dfrac{10}{3} \hat{j}+\dfrac{10}{3} \hat{k}$ or $\displaystyle \dfrac{5}{3} \hat{i}-\dfrac{10}{3} \hat{j}-\dfrac{10}{3} \hat{k}$
Alternative approach:
Let $\displaystyle \vec{a}=4 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=-2 \hat{i}+\hat{j}-2 \hat{k}$ and let required vector be $\displaystyle \vec{p}=x \hat{i}+y \hat{j}+z \hat{k}$
so, $\displaystyle 25=x^{2}+y^{2}+z^{2}, \quad 4 x-y+3 z=0, \quad-2 x+y-2 z=0$
on solving we get, $\displaystyle z=-2 x, y=z \Rightarrow \dfrac{100}{9}=z^{2} \Rightarrow z= \pm \dfrac{10}{3}$
$\displaystyle \therefore$ Required vector, $\displaystyle \vec{p}=\dfrac{-5}{3} \hat{i}+\dfrac{10}{3} \hat{j}+\dfrac{10}{3} \hat{k}$ or $\displaystyle \dfrac{5}{3} \hat{i}-\dfrac{10}{3} \hat{j}-\dfrac{10}{3} \hat{k}$
$\displaystyle \cos^{2}\left(\dfrac{\pi}{4}\right)+\cos^{2}\left(\dfrac{\pi}{3}\right)+\cos^{2} \theta=1 \Rightarrow \cos \theta=\dfrac{1}{2}$ (As, $\displaystyle \theta$ is acute)
$\displaystyle \Rightarrow \theta=\dfrac{\pi}{3}$ & Components of $\displaystyle \vec{a}=\dfrac{1}{\sqrt{2}}, \dfrac{1}{2}, \dfrac{1}{2}$
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