Question

Find the area of the parallelogram with vertices: P(0,0,0), Q(-4,3,-1), R(-4,2,-2), S(-8,5,-3).

Answer 1

Answer:

Step-by-step explanation:

Co-ordinate of point P(0,0,0)

Q (-4,3,-1)

R (-4,2,-2)

S (-8,5,-3)

$\vec{PQ}=$

$\vec{PS}=$

Area of Parallelogram $=|\vec{PQ}\times \vec{PS}|$

$Area=\begin{vmatrix}i &j &k \\ -4 &3 &-1 \\ -8 &5 &-3 \end{vmatrix}$

$A=|\hat{i}(-9+5)-\hat{j}(-12-8)+\hat{k}(-20+24)|$

$A=|-4\hat{i}+20\hat{j}+4\hat{k}|$

$A=\sqrt{(-4)^2+(20)^2+(4)^2}$

$A=\sqrt{432}$

$A=20.78\ unit^2$