Question

Let $\vec{a}=2\hat{i}-\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}+3\hat{k}$. Let $\vec{v}$ be the vector in the plane of $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\dfrac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to

• A. $\dfrac{\sqrt{35}}{2}$
• B. $\dfrac{\sqrt{21}}{2}$
• C. 7
• D. 13

Hint:

For projection problems, always simplify the projection expression before expanding vector components.

Answer 1

The Correct Option is C

Step 1: Use projection formula.
Length of projection of vector v on vector c is given by:
|v · c| / |c|

Given:
|v · c| / |c| = 1 / √14

Step 2: Compute magnitude of vector c.
|c| = √(4 + 1 + 9) = √14

Thus:
|v · c| = 1

Step 3: Express vector v in the plane of vectors a and b.
v = m a + n b

v = (2m + n)i + (−m + 3n)j + (−m − n)k

Step 4: Take dot product with vector c.
v · c = 7(m + n)
|7(m + n)| = 1

|m + n| = 1 / 7

Step 5: Find magnitude of vector v.
|v| = 7