Question

If $\vec{a}=\hat{i}-2\hat{j}+2\hat{k}$ and $\vec{b}=9\hat{i}+6\hat{j}-18\hat{k}$ are two vectors, then $\frac{\text{Projection of } \vec{b} \text{ on } \vec{a}}{\text{Projection of } \vec{a} \text{ on } \vec{b}} =$

• A. 21
• B. 7
• C. $7/3$
• D. 3

Hint:

The projection of vector $\vec{u}$ on $\vec{v}$ is a scalar quantity. The ratio of projections of $\vec{b}$ on $\vec{a}$ to $\vec{a}$ on $\vec{b}$ simplifies to the ratio of their magnitudes, $|\vec{b}|/|\vec{a}|$.

Answer 1

The Correct Option is B

The projection of a vector $\vec{u}$ on a vector $\vec{v}$ is given by the formula $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$.
Let $P_1$ be the projection of $\vec{b}$ on $\vec{a}$.
$P_1 = \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$.
Let $P_2$ be the projection of $\vec{a}$ on $\vec{b}$.
$P_2 = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
We need to find the ratio $\frac{P_1}{P_2}$.
$\frac{P_1}{P_2} = \frac{\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}}{\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}}$.
Since the dot product is commutative ($\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$), the dot product terms cancel out.
The ratio simplifies to $\frac{|\vec{b}|}{|\vec{a}|}$.
First, we calculate the magnitude of $\vec{a}$.
$|\vec{a}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.
Next, we calculate the magnitude of $\vec{b}$.
$|\vec{b}| = \sqrt{9^2 + 6^2 + (-18)^2} = \sqrt{81+36+324} = \sqrt{441} = 21$.
The required ratio is $\frac{21}{3} = 7$.