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Find the vector equation of a plane whi...

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector `3 hat i+5 hat j-6 hat k`.

A

`(3)/(sqrt(70))x+(5)/(sqrt(70))y-(6)/(sqrt(70))z=7`

B

`3x+5y-6z=7`

C

`3sqrt(70)x+5sqrt(70)y-6sqrt(70z)=7`

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
A
Given the normal vector is,
`n=3hat(i)+5hat(j)-6hat(k)`
`:." "hat(n)=(n)/(|n|)=(3hat(i)+5hat(j)-6hat(k))/(sqrt(3^(2)+^(2)+(-6)^(2)))`
`=(3hat(i)+5hat(j)-6hat(k))/(sqrt(70))`
`=(3)/(sqrt(70))hat(i)+(5)/(sqrt(70))hat(j)-(6)/(sqrt(70))hat(k)`
The equation of the plane in normal form is `r*hat(n)=d`
`:.(xhat(i)+yhat(j)+zhat(k))*((3)/(sqrt(70))hat(i)+(5)/(sqrt(70))hat(j)-(6)/(sqrt(70)))=7`
`rArr" "(3)/(sqrt(70))x+(5)/(sqrt(70))y-(6)/(sqrt(70))z=7`
This is the vectors equation of the required plane.
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