Question

The scalar components of a unit vector which is perpendicular to each of the vectors \( \hat{i} + 2\hat{j} - \hat{k} \) and \( 3\hat{i} - \hat{j} + 2\hat{k} \) are:

• A. \( \left( \frac{-3}{\sqrt{83}}, \frac{-5}{\sqrt{83}}, \frac{7}{\sqrt{83}} \right) \)
• B. \( (-3, -5, 7) \)
• C. \( \left( \frac{-3}{\sqrt{83}}, \frac{-5}{\sqrt{83}}, \frac{-7}{\sqrt{83}} \right) \)
• D. \( (3, -5, -7) \)

Hint:

To find a unit vector perpendicular to two vectors, compute the cross product of the vectors and then normalize the result by dividing each component by the magnitude of the cross product.

Answer 1

The Correct Option is C

We are asked to find the scalar components of a unit vector that is perpendicular to both of the given vectors: - \( \mathbf{A} = \hat{i} + 2\hat{j} - \hat{k} \) - \( \mathbf{B} = 3\hat{i} - \hat{j} + 2\hat{k} \) To find a vector that is perpendicular to both, we take the cross product of \( \mathbf{A} \) and \( \mathbf{B} \). Step 1: Compute the cross product \( \mathbf{A} \times \mathbf{B} \). The cross product of \( \mathbf{A} \) and \( \mathbf{B} \) is given by the determinant: \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & -1
3 & -1 & 2 \end{vmatrix} \] Expanding the determinant: \[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 2 & -1
-1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & -1
3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2
3 & -1 \end{vmatrix} \] Now, compute each 2x2 determinant: - For \( \hat{i} \): \( \begin{vmatrix} 2 & -1
-1 & 2 \end{vmatrix} = 2(B) - (-1)(-1) = 4 - 1 = 3 \) - For \( \hat{j} \): \( \begin{vmatrix} 1 & -1
3 & 2 \end{vmatrix} = 1(B) - (-1)(3) = 2 + 3 = 5 \) - For \( \hat{k} \): \( \begin{vmatrix} 1 & 2
3 & -1 \end{vmatrix} = 1(-1) - 2(3) = -1 - 6 = -7 \) Thus, the cross product is: \[ \mathbf{A} \times \mathbf{B} = 3\hat{i} - 5\hat{j} - 7\hat{k} \] Step 2: Normalize the cross product. To find the unit vector, we first compute the magnitude of the cross product: \[ |\mathbf{A} \times \mathbf{B}| = \sqrt{3^2 + (-5)^2 + (-7)^2} = \sqrt{9 + 25 + 49} = \sqrt{83} \] Now, divide the components of the cross product by the magnitude to obtain the unit vector: \[ \hat{v} = \frac{1}{\sqrt{83}} \left( 3\hat{i} - 5\hat{j} - 7\hat{k} \right) \] Thus, the scalar components of the unit vector are: \[ \left( \frac{-3}{\sqrt{83}}, \frac{-5}{\sqrt{83}}, \frac{-7}{\sqrt{83}} \right) \] Therefore, the correct answer is option (C)