Question

$\text{If } \lambda>0, \text{ let } \theta \text{ be the angle between the vectors }\vec{a} = \hat{i} + \lambda \hat{j} - 3 \hat{k} \text{ and } \vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}.\text{ If the vectors } \vec{a} + \vec{b} \text{ and } \vec{a} - \vec{b} \text{ are mutually perpendicular, then the value of } (14 \cos \theta)^2 \text{ is equal to.}$

• A. 25
• B. 20
• C. 45
• D. 40

Answer 1

Answer: (A)

25

Answer 2

Given that the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular, we have:

$(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0.$

Expanding this, we get:

$\vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 0.$

Calculating $\vec{a} \cdot \vec{a}$ and $\vec{b} \cdot \vec{b}$:

$\vec{a} \cdot \vec{a} = (1)^2 + (\lambda)^2 + (-3)^2 = 1 + \lambda^2 + 9 = \lambda^2 + 10.$

$\vec{b} \cdot \vec{b} = (3)^2 + (-1)^2 + (2)^2 = 9 + 1 + 4 = 14.$

Setting $\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b}$:

$\lambda^2 + 10 = 14.$

$\lambda^2 = 4.$

$\lambda = 2 \quad (\text{since } \lambda > 0).$

Now, calculate $\vec{a} \cdot \vec{b}$ to find $\cos \theta$:

$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5.$

The magnitudes are:

$|\vec{a}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}.$

$|\vec{b}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}.$

Thus,

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-5}{14}.$

Now, we need to find $(14 \cos \theta)^2$:

$(14 \cos \theta)^2 = \left(14 \times \frac{-5}{14}\right)^2 = (-5)^2 = 25.$