Question

show that 4 points whose position vectors are 6i-7j, 16i-19j-4k, 3j-6k, 2i-5j+10k are coplaner

Answer 1

Let A, B, C, D are the points whose position vectors are 6iˆ−7jˆ, 16iˆ−19j
−4kˆ, 3jˆ−6kˆ, 2iˆ−5jˆ+10kˆ
AB−→− = position vector of B − position vector of
A =(16iˆ−19jˆ−4kˆ) − ( 6iˆ−7jˆ) = 10iˆ−12jˆ−4kˆ
AC−→− = position vector of C − position vector of A = ( 3jˆ−6kˆ) − (6iˆ−7jˆ) = −6iˆ + 10jˆ − 6kˆ
AD−→− = position vector of D − position vector of
A =( 2iˆ−5jˆ+10kˆ) − (6iˆ−7jˆ) = −4Iˆ+2jˆ+10kˆ
We know that three vectors are coplanar if their scalar product is zero.
If, [AB−→− AC−→− AD−→−] = 0
Now,
[AB−→− AC−→− AD−→−]
=|10 -12 -4 |
|-6 10 -6 |
|-4 2 6 |
= 10 (100 + 12) + 12 (−60 − 24) − 4 (−12 + 40)
= 10 × 112 + 12 × (−84) − 4 × 28
= 1120 − 1008 − 112
= 112 − 1120=0
=Hence, the given points are coplanar.

Answer 2

Answer:

hope the attachment helps you to understand the method.......

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