Question

the value of p so that the vectors 2i+j+k and i+2j-3k and 3i+pj+5k are coplanar should be​

Answer 1

Answer:

ps value is82(2838377475754748

Answer 2

Given :  vectors 2i+j+k and i+2j-3k and 3i+pj+5k are coplanar  

To Find : Value of p

Solution:

2i+j+k  

i+2j-3k

3i+pj+5k

vectors are coplanar if their scalar triple product is zero.

=>

$\left|\begin{array}{ccc}2&1&1\\1&2&-3\\3&p&5\end{array}\right| =0$

2(10 + 3p) - 1( 5 + 9) + 1( p - 6) = 0

=> 20 + 6p - 14 + p - 6 = 0

=> 7p = 0

=> p = 0

Hence value of p is zero

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