JEE MATHEMATICS MCQ: DCS AND DRS

1 Easy

1 of 1

Understanding +4 | -1

The distance between two points

P

and

Q

d

and the length of their projections of

P Q

on the co-ordinate planes are

d_1, d_2, d_3

. Then

d_1^2+d_2^2+d_3^2=k d^2

where '

k

' is

View answer and explanation

Correct option: D

Option D (2) is correct because the sum of the squares of the projections is equal to 2 times the square of the distance between the points.

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