Question

In the triangle with vertices $A(1,-1,2),B(5,-6,2)$ and $C(1,3,-1)$ find the altitude $n=|BD|$.

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

Answer 1

The correct option is A $5$

Given $A(1,-1,2),B(5,-6,2),C(1,3,-1)$

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

the equation of line $AC$ is $\frac{x-1}{0}=\frac{y+1}{4}=\frac{z-2}{-3}$

So the point of line $AC$ will be in the form of $(1,4t-1,2-3t)$

Since $D$ lies on $AC$ , we have $h=1,k=4t-1,l=2-3t$

The drs of $BD$ are $h-5,k+6,l-2$ and the drs of $AC$ are $0,4,-3$

Since $BD$ and $AC$ are perpendicular, we have $4(k+6)-3(l-2)=0$

$\Rightarrow 4(4t+5)-3(-3t)=0$

$\Rightarrow t=-\frac{4}{5}$

So, coordinates of $D$ are $D(1,-\frac{21}{5},\frac{22}{5})$

So, the value of $n=\sqrt{16+\frac{81}{25}+\frac{144}{25}}=\sqrt{25}=5$