Question

If \( \vec{a} = \hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} - 5\hat{j} + 6\hat{k} \), then the magnitude of the projection of \( 2\vec{a} - \vec{b} \) on \( \vec{a} + \vec{b} \) is:

• A. \( \dfrac{11\sqrt{2}}{\sqrt{10}} \)
• B. \( \dfrac{22}{\sqrt{10}} \)
• C. \( \dfrac{22}{\sqrt{133}} \)
• D. \( \dfrac{22}{\sqrt{5}} \)

Hint:

For projection problems: \[ \text{Projection of } \vec{u} \text{ on } \vec{v} = \frac{\vec{u}\cdot\vec{v}}{|\vec{v}|} \] Always remember:

• First simplify vectors completely.

• Use dot product carefully.

• Take modulus if magnitude is asked.

Answer 1

The Correct Option is C

Step 1: Recall the projection formula.
The magnitude of the projection of $\vec u$ on $\vec v$ is $\dfrac{|\vec u\cdot\vec v|}{|\vec v|}$.
Step 2: Build $2\vec a-\vec b$.
With $\vec a=(1,-1,3)$ and $\vec b=(3,-5,6)$, $2\vec a=(2,-2,6)$, so $2\vec a-\vec b=(-1,3,0)$.
Step 3: Build $\vec a+\vec b$.
Adding gives $\vec a+\vec b=(4,-6,9)$.
Step 4: Take the dot product.
$(-1)(4)+(3)(-6)+(0)(9)=-4-18=-22$, so its size is $22$.
Step 5: Find the length of $\vec a+\vec b$.
$|\vec a+\vec b|=\sqrt{16+36+81}=\sqrt{133}$.
Step 6: Apply the formula.
The projection magnitude is $\dfrac{22}{\sqrt{133}}$. \[ \boxed{\frac{22}{\sqrt{133}}} \]