Question

Let \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \), \( \vec{b} = 2\hat{i} + \hat{j} - \hat{k} \), \( \vec{c} = \lambda \hat{i} + \hat{j} + \hat{k} \) and \( \vec{v} = \vec{a} \times \vec{b} \). If \( \vec{v} \cdot \vec{c} = 11 \) and the length of the projection of \( \vec{b} \) on \( \vec{c} \) is \( p \), then find the value of \( 9p^2 \).

• A. 12
• B. 4
• C. 9
• D. 6

Hint:

Scalar Triple Product $[\vec{a} \vec{b} \vec{c}] = (\vec{a} \times \vec{b}) \cdot \vec{c}$.

Answer 1

The Correct Option is A

$\vec{a} = (1, -2, 3)$, $\vec{b} = (2, 1, -1)$.
$\vec{v} = \vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix} = (-1, 7, 5)$.
$\vec{c} = (\lambda, 1, 1)$.
$\vec{v} \cdot \vec{c} = -\lambda + 7 + 5 = 12 - \lambda = 11 \implies \lambda = 1$.
$\vec{c} = (1, 1, 1)$. $|\vec{c}| = \sqrt{3}$.
Projection $p = \frac{\vec{b}\cdot\vec{c}}{|\vec{c}|} = \frac{2(1) + 1(1) - 1(1)}{\sqrt{3}} = \frac{2}{\sqrt{3}}$.
$9p^2 = 9(4/3) = 12$.