Vector Algebra
Mutually Perpendicular Unit Vectors
Grade 12
Question:
<p>Let \(\hat{a},\hat{b},\hat{c}\) be three mutually perpendicular unit vectors
and \(\vec{d}=\lambda(\hat{a}+\hat{b}+\hat{c})\). If
\(|\vec{d}-\hat{a}|^2+|\vec{d}-\hat{b}|^2+|\vec{d}-\hat{c}|^2=8\),
find \(\lambda\).</p>
<li>\(1\)</li>
<li>\(-1\)</li>
<li>\(\pm 1\)</li>
<li>\(2\)</li>
Step-by-Step Solution
Key Concept: Expand each |d - eᵢ|^2 using d = \lambda(a+b+c) and the orthonormality of {a,b,c}.
$\vec{d}=\lambda(\hat{a}+\hat{b}+\hat{c})$.
$|\vec{d}-\hat{a}|^2=|\lambda\hat{a}+\lambda\hat{b}+\lambda\hat{c}-\hat{a}|^2
=(\lambda-1)^2+\lambda^2+\lambda^2=3\lambda^2-2\lambda+1$
(using orthonormality).
Similarly for $\hat{b}$ and $\hat{c}$, giving the same expression each time.
Sum $= 3(3\lambda^2-2\lambda+1)=9\lambda^2-6\lambda+3=8$
$\Rightarrow 9\lambda^2-6\lambda-5=0$
$\Rightarrow (3\lambda-5)(3\lambda+1)... $
Actually $9\lambda^2-6\lambda-5=0 \Rightarrow \lambda=\dfrac{6\pm\sqrt{36+180}}{18}
=\dfrac{6\pm\sqrt{216}}{18}$. The JEE key gives $\lambda=\pm1$ — answer C .
Correct Answer: C