If the distance of the point (1, 1, 1) from the origin is half its distance from the plane x+y+z+k=0, then k is equal to:

Question

If the distance of the point (1, 1, 1) from the origin is half its distance from the plane x+y+z+k=0, then k is equal to: (A)  ± 3  (B)  ± 6  (C)  -3,9  (D)  3,-9

Tardigrade Admin · Accepted Answer

Distance of a point (1, 1, 1) from x+y+z+k=0 is | (1+1+1+k/√3) | According to question, | (3+k/√3) |= ± 2√3 ⇒ k=3,-9

Q. If the distance of the point $(1, 1, 1)$ from the origin is half its distance from the plane $x + y + z + k = 0,$ then $k$ is equal to:

$\pm 3$

$\pm 6$

$- 3, 9$

$3, - 9$

3, 9

Distance of a point (1, 1, 1) from $x + y + z + k = 0$
is $∣ ∣ \frac{1 + 1 + 1 + k}{3} ∣ ∣$
According to question, $∣ ∣ \frac{3 + k}{3} ∣ ∣ =$
$\pm 23$ $\Rightarrow$ $k = 3, - 9$

Q. If the distance of the point (1,1,1) from the origin is half its distance from the plane x+y+z+k=0, then k is equal to:

±3

±6

−3,9

3,−9