If a plane cuts off intercepts OA = a, OB = b | vedclass

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(A) The vertices of the triangle $ABC$ are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The vectors representing the sides are $\vec{AB} = (-a, b, 0)$ a

Vedclass · Accepted Answer

$\frac{1}{2}\sqrt{b^2c^2 + c^2a^2 + a^2b^2}$

Vedclass · Answer

(A) The vertices of the triangle $ABC$ are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The vectors representing the sides are $\vec{AB} = (-a, b, 0)$ and $\vec{AC} = (-a, 0, c)$.The area of the triangle $ABC$ is given by $\Delta = \frac{1}{2} |\vec{AB} 	imes \vec{AC}|$.Calculating the cross product: $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -a & b & 0 \ -a & 0 & c \end{vmatrix} = \hat{i}(bc) - \hat{j}(-ac) + \hat{k}(ab) = (bc, ac, ab)$.The magnitude is $|\vec{AB} 	imes \vec{AC}| = \sqrt{(bc)^2 + (ac)^2 + (ab)^2} = \sqrt{b^2c^2 + a^2c^2 + a^2b^2}$.Therefore,the area is $\Delta = \frac{1}{2} \sqrt{b^2c^2 + c^2a^2 + a^2b^2}$.

If a plane cuts off intercepts $OA = a, OB = b, OC = c$ from the coordinate axes,then the area of the triangle $ABC$ is:

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