The value of k such that (x-4/1)=(y-2/1) = (z-k/2) lies in the plane 2x - 4y + z = 7, is

Question

The value of k such that (x-4/1)=(y-2/1) = (z-k/2) lies in the plane 2x - 4y + z = 7, is (A) 7 (B) -7 (C) No real value (D) 4

Tardigrade Admin · Accepted Answer

Given equation of straight line (x-4/1)=(y-2/1) = (z-k/2) Since, the line lies in the plane 2x - 4y + z = 7. Hence, point (4,2, k) must satisfy the plane. ⇒ 8 - 8 + k = 7 ⇒ k = 7

Q. The value of $k$ such that $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2 x - 4 y + z = 7$ , is

7

-7

No real value

4

Given equation of straight line
$\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$
Since, the line lies in the plane $2 x - 4 y + z = 7$ .
Hence, point $(4, 2, k)$ must satisfy the plane.
$\Rightarrow 8 - 8 + k = 7 \Rightarrow k = 7$

Q. The value of k such that 1x−4​=1y−2​=2z−k​ lies in the plane 2x−4y+z=7, is

7

-7

No real value

4

Given equation of straight line 1x−4​=1y−2​=2z−k​ Since, the line lies in the plane 2x−4y+z=7. Hence, point (4,2,k) must satisfy the plane. ⇒8−8+k=7⇒k=7

Q. The value of $k$ such that $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2 x - 4 y + z = 7$ , is

Given equation of straight line
$\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$
Since, the line lies in the plane $2 x - 4 y + z = 7$ .
Hence, point $(4, 2, k)$ must satisfy the plane.
$\Rightarrow 8 - 8 + k = 7 \Rightarrow k = 7$