write the direction cosines of the normal to the plane 3x + 4y + 12z =52
Answers
Answered by
16
GIVEN,
3X+4Y+12Z=52
dividing both sides by √a2+b2+c2 we get,NOTE: a2 is a square,b2 is b sqare,c2 is c sqare
3/√9+16+144(X)+4/√9+16+144(Y)+12√9+16+144(z)=52/√9+16+144
=>3/13(x)+4/13(y)+12/13(z)=52/13=4
It is in the form form of lX + mY+nZ=P
where l,m,n are DC'S
therefore DC'S are (3/13,4/13,12/13)
Hope it helped u mate☺
3X+4Y+12Z=52
dividing both sides by √a2+b2+c2 we get,NOTE: a2 is a square,b2 is b sqare,c2 is c sqare
3/√9+16+144(X)+4/√9+16+144(Y)+12√9+16+144(z)=52/√9+16+144
=>3/13(x)+4/13(y)+12/13(z)=52/13=4
It is in the form form of lX + mY+nZ=P
where l,m,n are DC'S
therefore DC'S are (3/13,4/13,12/13)
Hope it helped u mate☺
Answered by
0
Step-by-step explanation:
hope it helps you,..........
Attachments:
Similar questions