Question

In a triangle with vertices $A(1,-1,2),B(5,-5,2)andC(1,3,-1),$ find the altitude $n=\left|BD\right|.$

• A. $5$
• B. $10$
• C. $5\sqrt{2}$
• D. $10\sqrt{2}$

Answer 1

The correct option is A

$5$

Explanation for the correct option:

Step 1: Finding the coordinates of $D$

Let the coordinates of $D$ be $(h,k,l).$

The equation of the line $AC$ can be written as: $\frac{x-1}{0}=\frac{y+1}{4}=\frac{z-2}{3}$.

Let $\frac{x-1}{0}=\frac{y+1}{4}=\frac{z-2}{3}=t$

Therefore, the point of the line $AC$ will be written in the form: $\left(1,4t-1,2-3t\right)$

Now, since $D$ lies on $AC$, it will also satisfy the form.

On equating $\left(1,4t-1,2-3t\right)$ and $(h,k,l).$ we get,

$h=1,k=4t-1,l=2-3t$

Step 2: Finding the altitude

$BD$ have $h-5,k+6,-2$ and $AC$ have $0,4,-3$

Since, $BD$ and $AC$ are perpendicular, we have:

$$\begin{gathered} 4\left(k+6\right)-3\left(l-2\right)=0 \\ \Rightarrow 4(4t+5)-3(-3t)=0 \\ \Rightarrow 16t+20+9t=0 \\ \Rightarrow 25t+20=0 \\ \Rightarrow t=\frac{-4}{5} \\ \therefore k=4t-1 \\ =4\left(\frac{-4}{5}\right)-1 \\ =\frac{-21}{5} \\ \therefore l=2-3t \\ =2-3\left(\frac{-4}{5}\right) \\ =\frac{22}{5} \end{gathered}$$

Therefore, the point $Dis\left(1,\frac{-21}{5},\frac{22}{5}\right)$

Hence,

$$\begin{gathered} n=\left|BD\right| \\ =\sqrt{1^2+{\left(\frac{21}{5}\right)}^2+{\left(\frac{22}{5}\right)}^2} \\ =\sqrt{25} \\ =5 \end{gathered}$$

Hence, the correct answer is option (A)