First slide

Cross product of vectors or vector product of vectors

Question

$Let A B C D be a tetrahedron such that the edges A B, A C and A D are$
$mutually perpendicular. Let the area of triangles ABC, ACD and ADB be 3, 4 and$
$5 sq. units respectively. Then the area of the triangle BCD is :$

Moderate

A

$5 \sqrt{2}$
B

5
C

$\frac{5}{\sqrt{2}}$
D

$\frac{5}{2}$

Solution

$\begin{array}{rcl} Area of ABCD & = & \frac{1}{2} | \vec{B C} X \vec{B D} | \\ = & \frac{1}{2} |(c \hat{j} - b \overset{\land}{i}) (d \hat{k} - b \overset{\land}{i})| \\ = & \frac{1}{2} | d c \overset{\land}{i} + b c \overset{\land}{k} + b d \overset{\land}{j} | \\ = & \frac{1}{2} \sqrt{b^{2} c^{2} + c^{2} d^{2} + d^{2} b^{2}} \\ \overset{}{} X ( \end{array}$
$Now 6 = b c; 8 = c d; 10 = b d b^{2} c^{2} + c^{2} d^{2} = d^{2} b^{2} = 200 Substituting the value in eqn. (i), A = \frac{1}{2} \sqrt{200} = 5 \sqrt{2}$

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