Question

Let $\overrightarrow{AB}=3\hat{i}+4\hat{k}$ and $\overrightarrow{AC}=5\hat{i}-2\hat{j}+4\hat{k}$. If the length of altitude from $A$ to the side $BC$ of $\triangle ABC$ is $P$ then the greatest integer less than $P^2$ is equal to:

• A. 20
• B. 18
• C. 19
• D. 21

Answer 1

Answer: (A)

20

Answer 2

1. Calculate the vector representing side BC: $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB} = (5\hat{i} - 2\hat{j} + 4\hat{k}) - (3\hat{i} + 4\hat{k}) = 2\hat{i} - 2\hat{j}$
2. Calculate the square of the length of side BC: $|\overrightarrow{BC}|^2 = (2)^2 + (-2)^2 + (0)^2 = 4 + 4 = 8$
3. Calculate the cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$: $\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 4 \\ 5 & -2 & 4 \end{vmatrix} = \hat{i}(0 - (-8)) - \hat{j}(12 - 20) + \hat{k}(-6 - 0) = 8\hat{i} + 8\hat{j} - 6\hat{k}$
4. Calculate the square of the magnitude of the cross product: $|\overrightarrow{AB} \times \overrightarrow{AC}|^2 = (8)^2 + (8)^2 + (-6)^2 = 64 + 64 + 36 = 164$
5. Relate the area of the triangle to the altitude: The area of $\triangle ABC$ is $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$. Also, Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} |\overrightarrow{BC}| P$. Equating these gives: $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} |\overrightarrow{BC}| P$ $|\overrightarrow{AB} \times \overrightarrow{AC}| = |\overrightarrow{BC}| P$ Squaring both sides: $|\overrightarrow{AB} \times \overrightarrow{AC}|^2 = |\overrightarrow{BC}|^2 P^2$
6. Calculate $P^2$: $P^2 = \frac{|\overrightarrow{AB} \times \overrightarrow{AC}|^2}{|\overrightarrow{BC}|^2} = \frac{164}{8} = \frac{41}{2} = 20.5$
7. Find the greatest integer less than $P^2$: The greatest integer less than $20.5$ is $\lfloor 20.5 \rfloor = 20$.