Vector Algebra
Cross product and projection
Grade 12
Question:
<p>If \(\vec{x} = 3\hat{i} - 6\hat{j} - \hat{k}\), \(\vec{y} = \hat{i} + 4\hat{j} - 3\hat{k}\) and \(\vec{z} = 3\hat{i} - 4\hat{j} - 12\hat{k}\), then the magnitude of the projection of \(\vec{x} \times \vec{y}\) on \(\vec{z}\) is</p>
<p>12</p>
<p>15</p>
<p>14</p>
<p>13</p>
Step-by-Step Solution
Key Concept: The projection of vector $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. For projection of $\vec{x} \times \vec{y}$ on $\vec{z}$, use the scalar triple product: $\frac{|(\vec{x} \times \vec{y}) \cdot \vec{z}|}{|\vec{z}|}$.
Step 1: Calculate $\vec{x} \times \vec{y}$ using the determinant: $\vec{x} \times \vec{y} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & -1 \\ 1 & 4 & -3 \end{vmatrix}$ $= \hat{i}(18 + 4) - \hat{j}(-9 + 1) + \hat{k}(12 + 6)$ $= 22\hat{i} + 8\hat{j} + 18\hat{k}$ Step 2: Calculate $(\vec{x} \times \vec{y}) \cdot \vec{z}$ (scalar triple product): $(22\hat{i} + 8\hat{j} + 18\hat{k}) \cdot (3\hat{i} - 4\hat{j} - 12\hat{k})$ $= 22(3) + 8(-4) + 18(-12)$ $= 66 - 32 - 216 = -182$ Step 3: Calculate $|\vec{z}|$: $|\vec{z}| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ Step 4: Magnitude of projection: $\text{Projection} = \left|\frac{(\vec{x} \times \vec{y}) \cdot \vec{z}}{|\vec{z}|}\right| = \left|\frac{-182}{13}\right| = \frac{182}{13} = 14$ ∴ Answer: 14
Correct Answer: C