Finding the Perpendicular Vector

Given points are $A \ (0,4,3)$, $B \ (0,0,0)$, and $C \ (3,0,4)$. We need to find a vector perpendicular to both line vectors $\vec{BA}$ and $\vec{BC}$.

Calculating Line Vectors

First, calculate the vectors $\vec{BA}$ and $\vec{BC}$:

• Vector $\vec{BA}$ is found by subtracting the coordinates of B from A: $ \vec{BA} = A - B = (0-0)\hat{i} + (4-0)\hat{j} + (3-0)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k} $
• Vector $\vec{BC}$ is found by subtracting the coordinates of B from C: $ \vec{BC} = C - B = (3-0)\hat{i} + (0-0)\hat{j} + (4-0)\hat{k} = 3\hat{i} + 0\hat{j} + 4\hat{k} $

Using the Cross Product

A vector perpendicular to two given vectors can be found using their cross product. We calculate $ \vec{BA} \times \vec{BC} $.

The cross product is calculated as follows:

$ \vec{BA} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & 3 \\ 3 & 0 & 4 \end{vmatrix} $

Expanding the determinant:

• For $ \hat{i} $: $ (4 \times 4) - (3 \times 0) = 16 - 0 = 16 $
• For $ \hat{j} $: $ -[(0 \times 4) - (3 \times 3)] = -[0 - 9] = -(-9) = 9 $
• For $ \hat{k} $: $ (0 \times 0) - (4 \times 3) = 0 - 12 = -12 $

Therefore, the resulting vector is: $ \vec{BA} \times \vec{BC} = 16\hat{i} + 9\hat{j} - 12\hat{k} $

Comparing with Options

The calculated vector $ 16\hat{i} + 9\hat{j} - 12\hat{k} $ matches Option 1.