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

The equation of the line parallel to x31=y+35=2z53 and passing through the point (1,3,5) in vector form, is:

A
r=(i+5j+3k)+t(i+3j+5k)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
r=(i+3j+5k)+t(i+5j+3k)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
r=(i+5j+32k)+t(i+3j+5k)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
r=(i+3j+5k)+t(i+5j+32k)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is C r=(i+3j+5k)+t(i+5j+32k)
We have to find the equation of the line parallel to x31=y+35=2z53 and passing through the point (1,3,5) vector form.

We know that the equation of the line passing through the point with position vector a and parallel to the vector b is r=a+tb

Consider x31=y+35=2z53

Rewriting we get x31=y+35=z5232

Thus vector representation is b=i+5j+32k

Since line passes through (1,3,5), a=i+3j+5k

Hence the required equation of the line is r=(i+3j+5k)+t(i+5j+32k)

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