The line perpendicular to the plane $2 x - y + 5 z = 4$ passing through the point $(- 1, 0, 1)$ is ?

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{x + 1}{2} = - y = \frac{z - 1}{5}$
$\begin{matrix} P o i n t A (- 1, 0, 1) a n d B (2, - 1, 5) S o, e q u a t i o n o f l i n e . \frac{x + 1}{2} = - y = \frac{z - 1}{5} O p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

Q. Equation of a line in the plane

π : 2 x - y + z - 4 = 0

which is perpendicular to the line

l

whose equation is

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

and which passes through the point of intersection of

l

and

π

Q. Equation of the line passing through the point

(2, 3, 1)

and parallel to the line of the intersection of the plane

x - 2 y - z + 5 = 0 and x + y + 3 z = 6

Perpendicular are drawn from points on the line $\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z}{3}$ to the plane x + y + z = 3. The feet of perpendiculars lie on the line

Q. The image of the line

\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}

in the plane

2 x - y + z + 3 = 0

is the line:

Q. On which of the following lines lies the point of intersection of the line,

\frac{x - 4}{2} = \frac{y - 5}{2} = \frac{z - 3}{1}

and the plane,

x + y + z = 2

Join BYJU'S Learning Program

The line perpendicular to the plane 2x−y+5z=4 passing through the point (−1,0,1) is ?

The correct option is C x+12=−y=z−15PointA(−1,0,1)andB(2,−1,5)So,equationofline.x+12=−y=z−15OptionCiscorrectanswer.

The line perpendicular to the plane $2 x - y + 5 z = 4$ passing through the point $(- 1, 0, 1)$ is ?

The correct option is C $\frac{x + 1}{2} = - y = \frac{z - 1}{5}$
$\begin{matrix} P o i n t A (- 1, 0, 1) a n d B (2, - 1, 5) S o, e q u a t i o n o f l i n e . \frac{x + 1}{2} = - y = \frac{z - 1}{5} O p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$