Question

Vectors $\vec{a}, \vec{b}, \vec{c}$ have magnitudes 2, 4, 4. Projection of $\vec{b}$ on $\vec{a}$ equals projection of $\vec{c}$ on $\vec{a}$ and $\vec{b} \perp \vec{c}$. Value of $|\vec{a}+\vec{b}-\vec{c}|$ is}

• A. 5
• B. 36
• C. 6
• D. 25

Hint:

$|\vec{u}|^{2} = \vec{u} \cdot \vec{u}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the magnitude of a linear combination of vectors using dot product properties derived from projection and perpendicularity.

Step 2: Key Formula or Approach:
1. Projection of $\vec{x}$ on $\vec{y} = \frac{\vec{x} \cdot \vec{y}}{|\vec{y}|}$.
2. Perpendicular vectors: $\vec{x} \cdot \vec{y} = 0$.
3. Magnitude squared: $|\vec{u}|^2 = \vec{u} \cdot \vec{u}$.

Step 3: Detailed Explanation:
Projection of $\vec{b}$ on $\vec{a} =$ Projection of $\vec{c}$ on $\vec{a} \Rightarrow \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|} = \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} \Rightarrow \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$.
$\vec{b} \perp \vec{c} \Rightarrow \vec{b} \cdot \vec{c} = 0$.
Consider $|\vec{a} + \vec{b} - \vec{c}|^2$: \[ = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b}) - 2(\vec{b} \cdot \vec{c}) - 2(\vec{a} \cdot \vec{c}) \] Since $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$, the terms $2(\vec{a} \cdot \vec{b})$ and $-2(\vec{a} \cdot \vec{c})$ cancel out.
Since $\vec{b} \cdot \vec{c} = 0$, that term is zero.
\[ = 2^2 + 4^2 + 4^2 + 0 - 0 - 0 \] \[ = 4 + 16 + 16 = 36 \] Taking square root: $|\vec{a} + \vec{b} - \vec{c}| = \sqrt{36} = 6$.

Step 4: Final Answer:
The magnitude is $6$.