JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 18)
If two vectors $$\overrightarrow P = \widehat i + 2m\widehat j + m\widehat k$$ and $$\overrightarrow Q = 4\widehat i - 2\widehat j + m\widehat k$$ are perpendicular to each other. Then, the value of m will be :
$$-1$$
3
1
2
Explanation
$\vec{P} \,\&\, \vec{Q}$ are perpendicular
$$ \begin{aligned} & \overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}}=0 \\\\ & (\hat{\mathrm{i}}+2 \mathrm{~m} \hat{\mathrm{j}}+\mathrm{m} \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\mathrm{m} \hat{\mathrm{k}})=0 \\\\ & \Rightarrow 4-4 \mathrm{~m}+\mathrm{m}^2=0 \\\\ & \Rightarrow(\mathrm{m}-2)^2=0 \Rightarrow \mathrm{m}=2 \end{aligned} $$
$$ \begin{aligned} & \overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}}=0 \\\\ & (\hat{\mathrm{i}}+2 \mathrm{~m} \hat{\mathrm{j}}+\mathrm{m} \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\mathrm{m} \hat{\mathrm{k}})=0 \\\\ & \Rightarrow 4-4 \mathrm{~m}+\mathrm{m}^2=0 \\\\ & \Rightarrow(\mathrm{m}-2)^2=0 \Rightarrow \mathrm{m}=2 \end{aligned} $$
Comments (0)
