MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Distance Formula

The value of ...

Question

The value of k for which the planes 3x - 6y - 2z = 7 and 2x + y - kz = 5 are perpendicular to each other, is
[MP PET 1992]

A

0

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

B

1

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

C

2

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

D

3

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

Open in App

Solution

The correct option is A
0

Planes are perpendicular, if
$6 - 6 + 2 k = 0 \Rightarrow k = 0$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The value of

k

for which the planes

3 x - 6 y - 2 z = 7

2 x + y - k z = 5

are perpendicular to each other, is

3 x - 6 y - 2 z = 7 and 2 x + y - λ z = 5

are perpendicular. Then the value of

λ

Q. Determine the value of λ for which the following planes are perpendicular to each other.
(i)

\vec{r} \cdot (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 7 and \vec{r} \cdot (λ \hat{i} + 2 \hat{j} - 7 \hat{k}) = 26

(ii) 2x − 4y + 3z = 5 and x + 2y + λz = 5

(iii) 3x − 6y − 2z = 7 and 2x + y − λz = 5

Q. The equation of plane passing through the point

(1, 1, 1)

and perpendicular to the planes

2 x + y - 2 z = 5

3 x - 6 y - 2 z = 7

Q. The equation of plane passing through the point

(1, 1, 1)

and perpendicular to the planes

2 x + y - 2 z = 5

3 x - 6 y - 2 z = 7

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Distance Formula

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Distance Formula

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon