The distance of the point (1,-5,9) from the plane x-y+z=5 measured along a straight line x=y=z is

Question

The distance of the point (1,-5,9) from the plane x-y+z=5 measured along a straight line x=y=z is (A) 3 √5 (B) 10 √3 (C) 5 √3 (D) 3 √10

Tardigrade Admin · Accepted Answer

Equation of P Q, is (x-1/1)=(x+5/1)=(z-9/1) x=λ+1 y=λ-5 z=λ+9 lies in Then plane x-y+z=5 λ+1-λ+5+λ+9=5 λ=-10 Q(-9,-15,-1) and P(1,-5,9) ⇒ |P Q|=√102+102+102=√300=10 √3

Q. The distance of the point $(1, - 5, 9)$ from the plane $x - y + z = 5$ measured along a straight line $x = y = z$ is

$35$

$103$

$53$

$310$

Equation of $PQ$ , is $\frac{x - 1}{1} = \frac{x + 5}{1} = \frac{z - 9}{1}$
$x = λ + 1 y = λ - 5 z = λ + 9$ lies in
Then plane $x - y + z = 5$
$λ + 1 - λ + 5 + λ + 9 = 5$
$λ = - 10$
$Q (- 9, - 15, - 1)$ and $P (1, - 5, 9)$
$\Rightarrow ∣ PQ ∣ = 1 0^{2} + 1 0^{2} + 1 0^{2} = 300 = 103$

Q. The distance of the point (1,−5,9) from the plane x−y+z=5 measured along a straight line x=y=z is

35​

103​

53​

310​