Question

If the position vectors of the vertices A, B, C of a triangle are $3\mathbf{i} + 4\mathbf{j} - \mathbf{k}$, $\mathbf{i} + 3\mathbf{j} + \mathbf{k}$, and $5(\mathbf{i} + \mathbf{j} + \mathbf{k})$ respectively, then the magnitude of the altitude drawn from A onto the side BC is

• A. $\frac{4\sqrt{5}}{3}$
• B. $\frac{5\sqrt{5}}{3}$
• C. $\frac{7\sqrt{5}}{3}$
• D. $\frac{8\sqrt{5}}{3}$

Hint:

Compute the area of a triangle using the cross product $\frac{1}{2} |\mathbf{AB} \times \mathbf{AC}|$, and find the altitude using $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$.

Answer 1

The Correct Option is A

Given vertices $A(3, 4, -1)$, $B(1, 3, 1)$, $C(5, 5, 5)$, find the altitude from $A$ to $BC$. The area of $\triangle ABC$ is: \[ \mathbf{AB} = \mathbf{B} - \mathbf{A} = (1 - 3, 3 - 4, 1 - (-1)) = -2\mathbf{i} - \mathbf{j} + 2\mathbf{k} \] \[ \mathbf{AC} = \mathbf{C} - \mathbf{A} = (5 - 3, 5 - 4, 5 - (-1)) = 2\mathbf{i} + \mathbf{j} + 6\mathbf{k} \] \[ \mathbf{AB} \times \mathbf{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & -1 & 2 \\ 2 & 1 & 6 \end{vmatrix} = \mathbf{i}(-6 - 2) - \mathbf{j}(-12 - 4) + \mathbf{k}(-2 - (-2)) = -8\mathbf{i} + 16\mathbf{j} \] \[ |\mathbf{AB} \times \mathbf{AC}| = \sqrt{(-8)^2 + 16^2} = \sqrt{64 + 256} = \sqrt{320} = 8\sqrt{5} \] \[ \text{Area} = \frac{1}{2} |\mathbf{AB} \times \mathbf{AC}| = \frac{1}{2} \cdot 8\sqrt{5} = 4\sqrt{5} \] The base $BC$ is: \[ \mathbf{BC} = \mathbf{C} - \mathbf{B} = (5 - 1, 5 - 3, 5 - 1) = 4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k} \] \[ |\mathbf{BC}| = \sqrt{4^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \] The altitude $h$ from $A$ satisfies: \[ \text{Area} = \frac{1}{2} \cdot |\mathbf{BC}| \cdot h \implies 4\sqrt{5} = \frac{1}{2} \cdot 6 \cdot h \implies 4\sqrt{5} = 3h \implies h = \frac{4\sqrt{5}}{3} \] Option (1) is correct. Options (2), (3), and (4) do not match.