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

Concept: The projection of a vector \( \vec{u} \) on another vector \( \vec{v} \) is given by: \[ \text{Projection magnitude} = \frac{|\vec{u}\cdot\vec{v}|}{|\vec{v}|} \] Thus, we first determine the vectors involved, compute their dot product carefully, and then divide by the magnitude of the vector on which projection is taken.

Step 1: Find the vector \( 2\vec{a} - \vec{b} \). Given: \[ \vec{a} = \hat{i} - \hat{j} + 3\hat{k} \] \[ \vec{b} = 3\hat{i} - 5\hat{j} + 6\hat{k} \] Now, \[ 2\vec{a} = 2\hat{i} - 2\hat{j} + 6\hat{k} \] Therefore, \[ 2\vec{a} - \vec{b} = (2-3)\hat{i} + (-2+5)\hat{j} + (6-6)\hat{k} \] \[ 2\vec{a} - \vec{b} = -\hat{i} + 3\hat{j} \] Hence, \[ 2\vec{a} - \vec{b} = (-1,3,0) \]

Step 2: Find the vector \( \vec{a} + \vec{b} \). \[ \vec{a} + \vec{b} = (1+3)\hat{i} + (-1-5)\hat{j} + (3+6)\hat{k} \] \[ \vec{a} + \vec{b} = 4\hat{i} - 6\hat{j} + 9\hat{k} \] Thus, \[ \vec{a} + \vec{b} = (4,-6,9) \]

Step 3: Compute the dot product. \[ (2\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b}) = (-1)(4) + (3)(-6) + (0)(9) \] \[ = -4 -18 \] \[ = -22 \] Therefore, \[ |(2\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b})| = 22 \]

Step 4: Find the magnitude of \( \vec{a}+\vec{b} \). \[ |\vec{a}+\vec{b}| = \sqrt{4^2+(-6)^2+9^2} \] \[ = \sqrt{16+36+81} \] \[ = \sqrt{133} \]

Step 5: Apply projection formula. \[ \text{Projection magnitude} = \frac{22}{\sqrt{133}} \] Hence, the required answer is: \[ \boxed{\dfrac{22}{\sqrt{133}}} \]