Vector Algebra
Position Vectors and Geometry
Grade None
Question:
<p>[JEE Main 2021] Let \(\vec{a},\vec{b},\vec{c}\) be three mutually perpendicular unit vectors. The angle \(\theta\) between each of them and the vector \(\vec{a}+\vec{b}+\vec{c}\) is</p>
<li>\(\cos^{-1}\dfrac{1}{\sqrt2}\)</li>
<li>\(\cos^{-1}\dfrac{1}{\sqrt3}\)</li>
<li>\(\tan^{-1}\sqrt2\)</li>
<li>\(\sin^{-1}\dfrac{1}{\sqrt3}\)</li>
Step-by-Step Solution
Key Concept: a \cdot (a+b+c) = |a|^2 = 1 (since a\perpb, a\perpc). |a+b+c| = \sqrt{3} (unit vectors mutually \perp). cos\theta = 1/\sqrt{3.}
Since $\vec{a},\vec{b},\vec{c}$ are mutually perpendicular unit vectors:
$\vec{a}\cdot(\vec{a}+\vec{b}+\vec{c})=|\vec{a}|^2+0+0=1$.
$|\vec{a}+\vec{b}+\vec{c}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2}=\sqrt3$ (cross terms are zero).
$\cos\theta=\dfrac{1}{1\cdot\sqrt3}=\dfrac{1}{\sqrt3}\Rightarrow\theta=\cos^{-1}\!\dfrac{1}{\sqrt3}=\tan^{-1}\sqrt2$.
Answer: (C) $\tan^{-1}\sqrt2$ (which equals $\cos^{-1}(1/\sqrt3)$).
Correct Answer: C