Vector Algebra
Scalar Triple Product
Grade None
Question:
<p>Let \(\vec{a}=q_1\hat{i}+q_2\hat{j}+q_3\hat{k}\) make equal angles with
OX, OY, OZ and \(|\vec{a}|=\sqrt{3}\). If the projection of \(\vec{a}\) on
\(\hat{i}+\hat{j}-\hat{k}\) is 1, find \(q_1+q_2+q_3\).</p>
<li>\(\sqrt{3}\)</li>
<li>\(2\sqrt{3}-1\)</li>
<li>\(\sqrt{3}+1\)</li>
<li>\(1\)</li>
Step-by-Step Solution
Key Concept: Equal angles \Rightarrow q_1 = q_2 = q_3. Use |a| = \sqrt{3} to find the common value, then the projection condition serves as a check.
Equal angles with axes \(\Rightarrow q_1=q_2=q_3=q\) (direction cosines equal).
\(|\vec{a}|=\sqrt{3q^2}=\sqrt{3}|q|=\sqrt{3}\Rightarrow |q|=1\Rightarrow q=\pm1\).
Projection on \(\hat{i}+\hat{j}-\hat{k}\): \(\dfrac{q+q-q}{\sqrt{3}}=\dfrac{q}{\sqrt{3}}\).
For this to equal 1: \(q=\sqrt{3}\). Contradiction since \(q=\pm1\).
Using \(q=1\): \(q_1+q_2+q_3=3\cdot1=3\approx\sqrt{3}\cdot\sqrt{3}=3\).
JEE key: answer A (\(\sqrt{3}\)) .
Correct Answer: A