wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

What is the distance of the plane 2x3y+4z=6 from the origin?


A

229

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

329

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

529

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

629

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

The correct option is D

629


We know that the equation of plane in normal vector form is r^n=d

Where ^n is the unit vector in the direction of normal.

Let r vector be equal to x^i+y^j+z^k

And ^n be equal to l^i+m^j+n^k

Where l2+m2+n2=1

So the equation of plane will be -

(x^i+y^j+z^k).(l^i+m^j+n^k)=d

Or lx+my+nz=d

Here 'd' is the distance from origin.

The equation we are given in the question doesn’t have l,m and n as the coefficients of x,y and z, Since 22+(3)2+(4)21

So, we’ll find appropriate l,m,n first.

Here, We can consider 2^i3^j+4^k as a vector. And now we have to convert it to a unit vector.

We know that n=|n|.^n

Thus, the unit vector here will be =2i3j+4k|2i3j+4k|

=2i3j+4k29

So, l=229, m=329,n=429

To have l,m,n in the equation given we’ll divide both L.H.S and R.H.S by 29

So, we’ll have 2x3y+4z29=629

Or 229x329y+429z=629

Since the coefficients of x,y and z are such that they represent the direction cosines. The constant term on the R.H.S will be equal to the distance of the plane from origin.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Equation of a Plane - Normal Form
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon