JEE MATHEMATICS MCQ: DCS AND DRS

1 Easy

1 of 1

Understanding +4 | -1

Let

A_{x y}, A_{y z}, A_{z x}

be the area of the projection of a plane area

A

on the

x y, y z, z x

plane respectively Then

A^2=

View answer and explanation

Correct option: A

A_{x y}^2+A_{y z}^2+A_{z x}^2

The correct answer is A (

A_{xy}^2+A_{yz}^2+A_{zx}^2

) because it represents the relationship between the area of a plane and its projections onto the coordinate planes, using direction cosines.

More Questions from DCS AND DRS

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\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z

\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7},

4 \lambda+9 \mu

\frac{x}{2}=\frac{y}{2}=\frac{z}{1}

\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}

4 {~m}

2 {~kg}

10 {~kg}

45^{\circ}

60^{\circ}

X

Y

Z

\mathrm{AB}

-2 / \sqrt{17}

3 / \sqrt{17},-2 / \sqrt{17}

\mathrm{AB}=\sqrt{17}

\mathrm{A}

(3,-6,10)

\mathrm{B}

A B

45^{\circ}

120^{\circ}

X

Y

A B

\theta

Z

\theta