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Mathematics · 2025 · 2 marks
CBSE 2025 · Region 5 · Set 1 · Q24
(a) Find a vector of magnitude 5 which is perpendicular to both the vectors 3 i -2 j + k and 4 i +3 j -2 k.
Find a vector of magnitude $\displaystyle 5$ which is perpendicular to both the vectors $\displaystyle 3 \hat{i}-2 \hat{j}+\hat{k}$ and $\displaystyle 4 \hat{i}+3 \hat{j}-2 \hat{k}$.
Let $\displaystyle \vec{\mathrm{a}}, \vec{\mathrm{b}}$ and $\displaystyle \vec{\mathrm{c}}$ be three vectors such that $\displaystyle \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=\vec{\mathrm{a}} \cdot \vec{\mathrm{c}}$ and $\displaystyle \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{a}} \neq 0$. Show that $\displaystyle \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}$.
Marking-scheme solution
Let $\displaystyle \overrightarrow{\mathbf{a}}=\mathbf{3 i} \boldsymbol{-} \mathbf{2 \hat { \jmath }}+\mathbf{\hat { k }}, \quad \overrightarrow{\mathbf{b}}=\mathbf{4 \hat { i }}+\mathbf{3 \hat { j }}-\mathbf{2 \hat { \mathbf { k } }}$
$\displaystyle \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -2 & 1 \\ 4 & 3 & -2 \end{array}\right|=\hat{\mathbf{i}}+10 \hat{\mathbf{j}}+17 \hat{\mathbf{k}}$
$\displaystyle |\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=\sqrt{\mathbf{1}^{\mathbf{2}}+\mathbf{1 0}^{\mathbf{2}}+\mathbf{1 7}^{\mathbf{2}}}=\sqrt{\mathbf{3 9 0}}$
Unit vector $\displaystyle \widehat{\mathbf{n}}=\frac{\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|}=\frac{\mathbf{1}}{\sqrt{\mathbf{3 9 0}}}(\hat{\mathbf{i}}+\mathbf{1 0} \hat{\mathbf{j}}+\mathbf{1 7} \hat{\mathbf{k}})$
∴ Required vector $\displaystyle =\frac{\mathbf{5}}{\sqrt{\mathbf{3 9 0}}}(\hat{\mathbf{i}}+\mathbf{1 0} \hat{\mathbf{j}}+\mathbf{1 7} \hat{\mathbf{k}})$
$\displaystyle \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=\vec{\mathrm{a}} \cdot \vec{\mathrm{c}} \Rightarrow \vec{\mathrm{a}} \cdot(\vec{\mathrm{b}}-\vec{\mathrm{c}})=0$
⇒ either $\displaystyle \vec{\mathrm{b}}=\vec{\mathrm{c}}$ or $\displaystyle \vec{\mathrm{a}} \perp(\vec{\mathrm{b}}-\vec{\mathrm{c}})$, since $\displaystyle \vec{\mathrm{a}} \neq 0$
Also, $\displaystyle \vec{\mathrm{a}} \times \vec{\mathrm{b}}=\vec{\mathrm{a}} \times \vec{\mathrm{c}} \Rightarrow \vec{\mathrm{a}} \times(\vec{\mathrm{b}}-\vec{\mathrm{c}})=0$
⇒ either $\displaystyle \vec{\mathrm{b}}=\vec{\mathrm{c}}$ or $\displaystyle \vec{\mathrm{a}} \|(\vec{\mathrm{b}}-\vec{\mathrm{c}})$, since $\displaystyle \vec{\mathrm{a}} \neq 0$
Since vectors $\displaystyle \overrightarrow{\boldsymbol{\mathrm{a}}}$ and $\displaystyle (\overrightarrow{\boldsymbol{\mathrm{b}}}-\overrightarrow{\boldsymbol{\mathrm{c}}})$ cannot be $\displaystyle \|$ and ⟂ simultaneously
Hence $\displaystyle \overrightarrow{\boldsymbol{\mathrm{b}}}=\overrightarrow{\boldsymbol{\mathrm{c}}}$
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