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: Concept
Projection of $\vec{u}$ on $\vec{v}$ is $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$.

Step 2: Analysis
$\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|} = \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} \implies \vec{b} \cdot \vec{a} = \vec{c} \cdot \vec{a}$.
Also, $\vec{b} \cdot \vec{c} = 0$ (perpendicular).

Step 3: Calculation
$|\vec{a}+\vec{b}-\vec{c}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} + |\vec{c}|^{2} + 2\vec{a}\cdot\vec{b} - 2\vec{b}\cdot\vec{c} - 2\vec{a}\cdot\vec{c}$
$= 2^{2} + 4^{2} + 4^{2} + 2(\vec{a}\cdot\vec{b} - \vec{a}\cdot\vec{c}) - 2(0)$
$= 4 + 16 + 16 + 2(0) = 36$.
$|\vec{a}+\vec{b}-\vec{c}| = \sqrt{36} = 6$.

Step 4: Conclusion
Hence, the value is 6. Final Answer: (C)