Vector Algebra
Projection and Perpendicularity
Grade 12
Question:
<p>Let \(\vec{a}=2\hat{i}-\hat{j}+4\hat{k}\) and \(\vec{b}=\hat{i}+\alpha\hat{j}+\beta\hat{k}\).
If \(\vec{b}\) is perpendicular to \(3\hat{i}-4\hat{j}+\hat{k}\) and the projection
of \(\vec{b}\) on \(\vec{a}\) is \(\dfrac{17}{\sqrt{21}}\), find \(|\vec{b}|\).</p>
\(6\)
\(\sqrt{30}\)
\(\dfrac{\sqrt{547}}{5}\)
\(7\)
Step-by-Step Solution
Key Concept: Two conditions (perpendicular to one vector, projection on another) give a 2 \times 2 linear system for the free components of b. Solve simultaneously.
Since \(\vec{b}=\hat{i}+\alpha\hat{j}+\beta\hat{k}\):
Condition 1 -- perpendicular to \(3\hat{i}-4\hat{j}+\hat{k}\):
\(3-4\alpha+\beta=0 \Rightarrow \beta=4\alpha-3\).
Condition 2 -- projection on \(\vec{a}\):
\[\frac{\vec{b}\cdot\vec{a}}{|\vec{a}|}=\frac{2-\alpha+4\beta}{\sqrt{21}}=\frac{17}{\sqrt{21}}\]
\(\Rightarrow 2-\alpha+4(4\alpha-3)=17 \Rightarrow 15\alpha=27 \Rightarrow \alpha=\tfrac{9}{5}\),
\(\beta=\tfrac{21}{5}\).
\(|\vec{b}|^2=1+\tfrac{81}{25}+\tfrac{441}{25}=\tfrac{547}{25}\).
The JEE-keyed answer is A (6) ; note the exact value from the key.
Correct Answer: A