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>
<li>\(6\)</li>
<li>\(\sqrt{30}\)</li>
<li>\(\dfrac{\sqrt{547}}{5}\)</li>
<li>\(7\)</li>
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