Vector Algebra
Cross product and perpendicular vectors
Grade 12
Question:
<p>The unit vector perpendicular to the vectors <span style='font-family:monospace'>a = 6<strong>i</strong> + 2<strong>j</strong> + 3<strong>k</strong></span> and <span style='font-family:monospace'>b = 3<strong>i</strong> - 6<strong>j</strong> - 2<strong>k</strong></span> is</p>
<p>(a) \(\frac{2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}}{7}\)</p>
<p>(b) \(\frac{2\mathbf{i} - 3\mathbf{j} - 6\mathbf{k}}{7}\)</p>
<p>(c) \(\frac{2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}}{7}\)</p>
<p>(d) \(\frac{2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}}{7}\)</p>
Step-by-Step Solution
Key Concept: A vector perpendicular to two given vectors is found using their cross product. The unit vector is obtained by dividing by the magnitude.
Step 1: Compute the cross product \(\mathbf{a} \times \mathbf{b}\): \[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & 2 & 3 \\ 3 & -6 & -2 \end{vmatrix}\] \(= \mathbf{i}(2(-2) - 3(-6)) - \mathbf{j}(6(-2) - 3(3)) + \mathbf{k}(6(-6) - 2(3))\) \(= \mathbf{i}(-4 + 18) - \mathbf{j}(-12 - 9) + \mathbf{k}(-36 - 6)\) \(= 14\mathbf{i} + 21\mathbf{j} - 42\mathbf{k}\) Step 2: Find magnitude: \(|\mathbf{a} \times \mathbf{b}| = \sqrt{196 + 441 + 1764} = \sqrt{2401} = 49\) Step 3: Unit vector: \(\hat{n} = \frac{14\mathbf{i} + 21\mathbf{j} - 42\mathbf{k}}{49} = \frac{2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}}{7}\) ∴ Answer is (c).
Correct Answer: c