Q.

The position vector of a point at a distance of  311 units from i^j^+2k^ on a line passing through the points i^j^+2k^ and 3i^+j^+k^ is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

-8i^-4j^k^

b

10i^2j^5k^

c

8i^+4j^+k^

d

10i^+2j^5k^

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

SOLUTION The equation of a line passing through the points

A(i^j^+2k^) and B(3i^+j^+k^) is given by 

r=(i^j^+2k^)+λ(3i^+j^+k^).

The position vector of a variable point P on the line, is 

(i^j^+2k^)+λ(3i^+j^+k^).

     AP=λ(3i^+j^+k^)|AP|=|λ|11 

Now, AP =311|λ|11=311,λ=±3

Thus, the position vectors of P are 

   10i^+2j^+5k^  and 8i^4j^k^.

Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
The position vector of a point at a distance of  311 units from i^−j^+2k^ on a line passing through the points i^−j^+2k^ and 3i^+j^+k^ is