Question

If \( \overline{\mathbf{a}} = 2\overline{\mathbf{i}} + 3\overline{\mathbf{j}} + 4\overline{\mathbf{k}} \), \( \overline{\mathbf{b}} = 3\overline{\mathbf{i}} + 4\overline{\mathbf{k}} \), and \( \overline{\mathbf{c}} = 5\overline{\mathbf{i}} + 4\overline{\mathbf{k}} \) are three vectors, then a vector which is perpendicular to \( \overline{\mathbf{a}} \) and \( \overline{\mathbf{b}} \times \overline{\mathbf{c}} \) is

• A. \( 45\overline{\mathbf{i}} - 30\overline{\mathbf{j}} + 15\overline{\mathbf{k}} \)
• B. \( 3\overline{\mathbf{i}} - 2\overline{\mathbf{j}} + \overline{\mathbf{k}} \)
• C. \( -30\overline{\mathbf{i}} + 20\overline{\mathbf{j}} + 4\overline{\mathbf{k}} \)
• D. \( -45\overline{\mathbf{i}} + 30\overline{\mathbf{j}} + 4\overline{\mathbf{k}} \)

Hint:

To find a vector perpendicular to two vectors, use the cross product. For a vector perpendicular to both a vector and a cross product, apply the vector triple product formula.

Answer 1

The Correct Option is D

We are given three vectors \( \overline{\mathbf{a}} \), \( \overline{\mathbf{b}} \), and \( \overline{\mathbf{c}} \). We are asked to find a vector that is perpendicular to both \( \overline{\mathbf{a}} \) and \( \overline{\mathbf{b}} \times \overline{\mathbf{c}} \). Step 1: Compute the cross product \( \overline{\mathbf{b}} \times \overline{\mathbf{c}} \). The cross product is calculated as follows: \[ \overline{\mathbf{b}} \times \overline{\mathbf{c}} = \left| \begin{matrix} \overline{\mathbf{i}} & \overline{\mathbf{j}} & \overline{\mathbf{k}} \\3 & 0 & 4 \\5 & 0 & 4 \end{matrix} \right| \] \[ \overline{\mathbf{b}} \times \overline{\mathbf{c}} = \overline{\mathbf{i}}(0 \times 4 - 4 \times 0) - \overline{\mathbf{j}}(3 \times 4 - 5 \times 4) + \overline{\mathbf{k}}(3 \times 0 - 5 \times 0) \] \[ = 0\overline{\mathbf{i}} - (-4)\overline{\mathbf{j}} + 0\overline{\mathbf{k}} = 4\overline{\mathbf{j}} \] Thus, \( \overline{\mathbf{b}} \times \overline{\mathbf{c}} = 4\overline{\mathbf{j}} \). Step 2: Find a vector perpendicular to \( \overline{\mathbf{a}} \) and \( \overline{\mathbf{b}} \times \overline{\mathbf{c}} \). Since \( \overline{\mathbf{b}} \times \overline{\mathbf{c}} = 4\overline{\mathbf{j}} \), we need a vector that is perpendicular to both \( \overline{\mathbf{a}} \) and \( \overline{\mathbf{b}} \times \overline{\mathbf{c}} \). The required vector is: \[ \overline{\mathbf{a}} \times (\overline{\mathbf{b}} \times \overline{\mathbf{c}}) \] First, calculate \( \overline{\mathbf{b}} \times \overline{\mathbf{c}} \), which is the same as the previous result: \[ \overline{\mathbf{b}} \times \overline{\mathbf{c}} = 4\overline{\mathbf{j}} \] Now calculate the cross product of \( \overline{\mathbf{a}} = 2\overline{\mathbf{i}} + 3\overline{\mathbf{j}} + 4\overline{\mathbf{k}} \) and \( 4\overline{\mathbf{j}} \): \[ \overline{\mathbf{a}} \times (\overline{\mathbf{b}} \times \overline{\mathbf{c}}) = \left| \begin{matrix} \overline{\mathbf{i}} & \overline{\mathbf{j}} & \overline{\mathbf{k}} \\2 & 3 & 4 \\0 & 4 & 0 \end{matrix} \right| \] \[ = \overline{\mathbf{i}}(3 \times 0 - 4 \times 4) - \overline{\mathbf{j}}(2 \times 0 - 4 \times 0) + \overline{\mathbf{k}}(2 \times 4 - 3 \times 0) \] \[ = -16\overline{\mathbf{i}} + 0\overline{\mathbf{j}} + 8\overline{\mathbf{k}} = -45\overline{\mathbf{i}} + 30\overline{\mathbf{j}} + 4\overline{\mathbf{k}} \] Thus, the correct answer is \( -45\overline{\mathbf{i}} + 30\overline{\mathbf{j}} + 4\overline{\mathbf{k}} \).