Finding the area of $\triangle ABC$ with $B$ and $C$ lying on the line $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$

Question

The vertices $B$ and $C$ of a $\triangle ABC$ lie on the line, $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$ such that $BC = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1, –1, 2)$, is: $5 \sqrt{17} /\sqrt{34}/ 6 / 2\sqrt{34}?$

My attempt: Let $B$ be $(3k+2,1,4k)$ and $C$ be $(3l+2,1,4l)$. So, using distance formula between $B$ and $C$, I get $k-l=1$. Now, using determinant, where the first row is coordinates of $A$, the second is of $B$, and the third $C$, I get area to be $5$. But the answer is given as $\sqrt{34}$. What is wrong in my method?

Answer 1

The problem with your method is that you're not measuring the area of triangle $\triangle ABC$, but the volume of the parallelepiped with vertices $0, A, B, C, A + B, A + C, B + C, A + B + C$. This is what the $3 \times 3$ determinant measures.

Instead, try computing $$\frac{1}{2}\|(B - A) \times (C - A)\|.$$ The vectors $B - A$ and $C - A$ are sides of the triangle, and the length of their cross product is the area of the parallelogram formed by these vectors. Half of this quantity will be the area of $\triangle ABC$.

Answer 2

Let $AD$ be an altitude of the triangle.

Thus, $$D(2+3t,1,4t)$$ and since $$\vec{AD}\cdot\vec{(3,0,4)}=0,$$ we obtain: $$(1+3t,2,4t-2)(3,0,4)=0,$$ which gives $$t=\frac{1}{5}.$$ Can you end it now?