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$\quad$ Let $\overrightarrow{\mathbf{a}}=2 \overrightarrow{\mathbf{j}}-3 \overrightarrow{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=-3 \overrightarrow{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$. Let $\hat{\mathbf{n}}$ be
a unit vector such that $\overrightarrow{\mathbf{a}} \hat{\mathbf{n}}=\overrightarrow{\mathbf{b}} \hat{\mathbf{n}}=0$. What is the value of $\overrightarrow{\mathbf{c}} . \hat{\mathbf{n}} ?$
Options:
a unit vector such that $\overrightarrow{\mathbf{a}} \hat{\mathbf{n}}=\overrightarrow{\mathbf{b}} \hat{\mathbf{n}}=0$. What is the value of $\overrightarrow{\mathbf{c}} . \hat{\mathbf{n}} ?$
Solution:
2463 Upvotes
Verified Answer
The correct answer is:
$-3$
As given $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
and $\overrightarrow{\mathrm{c}}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
Let $\overrightarrow{\mathrm{n}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$
Since $\vec{a}$ and $\hat{n}$ are perpendicular to each other. $\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{n}}=0 \Rightarrow(2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=0$
$\Rightarrow 2 \mathrm{y}-3 \mathrm{z}=0$
and $\overrightarrow{\mathrm{b}} \cdot \hat{\mathrm{n}}=0$
$\Rightarrow(\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=0$
$\Rightarrow \mathrm{y}+3 \mathrm{z}=0$
$\ldots(2)$
On solving Eqs. (1) and (2) $y=z=0$
Since $\hat{n}$ is a unit vector, $\sqrt{x^{2}+y^{2}+z^{2}}=1 \Rightarrow x=1 \quad[$ since, $y=z=0]$
hence, $\hat{n}=\hat{i}$
This gives, $\overrightarrow{\text { c. }} \hat{\text { n }}=(-3 \hat{i}+3 \hat{j}+\hat{k}) \cdot(\hat{\mathrm{i}})=-3$
and $\overrightarrow{\mathrm{c}}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
Let $\overrightarrow{\mathrm{n}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$
Since $\vec{a}$ and $\hat{n}$ are perpendicular to each other. $\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{n}}=0 \Rightarrow(2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=0$
$\Rightarrow 2 \mathrm{y}-3 \mathrm{z}=0$
and $\overrightarrow{\mathrm{b}} \cdot \hat{\mathrm{n}}=0$
$\Rightarrow(\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=0$
$\Rightarrow \mathrm{y}+3 \mathrm{z}=0$
$\ldots(2)$
On solving Eqs. (1) and (2) $y=z=0$
Since $\hat{n}$ is a unit vector, $\sqrt{x^{2}+y^{2}+z^{2}}=1 \Rightarrow x=1 \quad[$ since, $y=z=0]$
hence, $\hat{n}=\hat{i}$
This gives, $\overrightarrow{\text { c. }} \hat{\text { n }}=(-3 \hat{i}+3 \hat{j}+\hat{k}) \cdot(\hat{\mathrm{i}})=-3$
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