Question

Let $ \vec{a} = \hat{i} + \hat{j} + \hat{k} $, $ \vec{b} = 3\hat{i} + 2\hat{j} - \hat{k} $, $ \vec{c} = \lambda \hat{j} + \mu \hat{k} $ and $ \hat{d} $ be a unit vector such that $ \vec{a} \times \hat{d} = \vec{b} \times \hat{d} $ and $ \vec{c} \cdot \hat{d} = 1 $. If $ \vec{c} $ is perpendicular to $ \vec{a} $, then $ |3\lambda \hat{d} + \mu \vec{c}|^2 $ is equal to ____.

Hint:

Use the given vector relations to find the vectors \( \hat{d} \) and \( \vec{c} \). Remember that if \( \vec{a} \times \vec{b} = 0 \), then \( \vec{a} \) and \( \vec{b} \) are parallel.

Answer 1

Correct Answer: 5

Given \( \vec{a} \times \hat{d} = \vec{b} \times \hat{d} \) \[ \vec{a} \times \hat{d} - \vec{b} \times \hat{d} = 0 \] \[ (\vec{a} - \vec{b}) \times \hat{d} = 0 \] This means \( \vec{a} - \vec{b} \) is parallel to \( \hat{d} \). \[ \vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (3\hat{i} + 2\hat{j} - \hat{k}) = -2\hat{i} - \hat{j} + 2\hat{k} \] Let \[ \hat{d} = \frac{\vec{a} - \vec{b}}{|\vec{a} - \vec{b}|} = \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{(-2)^2 + (-1)^2 + 2^2}} = \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3} \] Given \( \vec{c} \cdot \hat{d} = 1 \): \[ (\lambda \hat{j} + \mu \hat{k}) \cdot \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3} = 1 \] \[ \frac{-\lambda + 2\mu}{3} = 1 \] \[ -\lambda + 2\mu = 3 \tag{1} \] Given \( \vec{c} \) is perpendicular to \( \vec{a} \), so \( \vec{c} \cdot \vec{a} = 0 \): \[ (\lambda \hat{j} + \mu \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0 \] \[ \lambda + \mu = 0 \Rightarrow \mu = -\lambda \tag{2} \] Substituting (2) in (1), we get: \[ -\lambda - 2\lambda = 3 \] \[ -3\lambda = 3 \Rightarrow \lambda = -1, \quad \mu = 1 \] So, \[ \vec{c} = -\hat{j} + \hat{k} \] Now, \[ 3\lambda \hat{d} + \mu \vec{c} = 3(-1) \cdot \frac{-2\hat{i} - \hat{j} + 2\hat{k}}{3} + (-\hat{j} + \hat{k}) \] \[ = 2\hat{i} + \hat{j} - 2\hat{k} - \hat{j} + \hat{k} = 2\hat{i} - \hat{k} \] \[ |3\lambda \hat{d} + \mu \vec{c}|^2 = |2\hat{i} - \hat{k}|^2 = 2^2 + (-1)^2 = 4 + 1 = 5 \]

Answer 2

Given vectors: \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = 3\hat{i} + 2\hat{j} - \hat{k} \), and \( \vec{c} = \lambda \hat{j} + \mu \hat{k} \). The unit vector \( \hat{d} \) satisfies \( \vec{a} \times \hat{d} = \vec{b} \times \hat{d} \) and \( \vec{c} \cdot \hat{d} = 1 \). Also, \( \vec{c} \) is perpendicular to \( \vec{a} \), so:

\( \vec{a} \cdot \vec{c} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\lambda \hat{j} + \mu \hat{k}) = \lambda + \mu = 0 \Rightarrow \mu = -\lambda \)

Considering \( \vec{a} \times \hat{d} = \vec{b} \times \hat{d} \), we equate cross products. For \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = 3\hat{i} + 2\hat{j} - \hat{k} \), calculate cross products:

\( \vec{a} \times \hat{d} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ d_1 & d_2 & d_3 \end{vmatrix} = (d_3 - d_2)\hat{i} + (d_1 - d_3)\hat{j} + (d_2 - d_1)\hat{k} \)

\( \vec{b} \times \hat{d} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ d_1 & d_2 & d_3 \end{vmatrix} = (-2d_3 - d_2)\hat{i} + (d_3 + 3d_1)\hat{j} + (2d_2 - 3d_1)\hat{k} \)

Equating coefficients of \( \hat{i}, \hat{j}, \hat{k} \) from both:

• \( d_3 - d_2 = -2d_3 - d_2 \Rightarrow 3d_3 = 0 \Rightarrow d_3 = 0 \)
• \( d_1 - d_3 = d_3 + 3d_1 \Rightarrow -2d_1 = 0 \Rightarrow d_1 = 0 \)
• \( d_2 - d_1 = 2d_2 - 3d_1 \Rightarrow d_2 = 0 \)

Since \( \hat{d} \) is a unit vector, assume \( \hat{d} = \hat{j} \). Hence, \( \vec{c} \cdot \hat{d} = 1 \Rightarrow \lambda = 1 \). Thus, \( \mu = -\lambda = -1 \). Given \( \vec{c} = \hat{j} - \hat{k} \), compute \( |3\lambda \hat{d} + \mu \vec{c}|^2 \):

\( \vec{x} = 3\lambda \hat{d} + \mu \vec{c} = 3\hat{j} - (\hat{j} - \hat{k}) = 2\hat{j} + \hat{k} \)

Calculate magnitude squared:

\( |2\hat{j} + \hat{k}|^2 = (2^2 + 1^2) = 4 + 1 = 5 \)

Thus, \( |3\lambda \hat{d} + \mu \vec{c}|^2 = 5 \), which fits the given range [5,5].