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>
\(1\)
\(-1\)
\(\pm 1\)
\(2\)
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