Question

15
Prove that the vectors A=2i-j+k,
b=i-3j-5k and c=3i-4j-4k
coplanar.
​

Answer 1

Answer:

Step-by-step explanation:

Given:

$\sf \overrightarrow{A}=2\hat{i}-\hat{j}+\hat{k}$

$\sf \overrightarrow{B}=\hat{i}-3\hat{j}-5\hat{k}$

$\sf \overrightarrow{C}=3\hat{i}-4\hat{j}-4\hat{k}$

To Prove:

The vectors are co planar

Proof:

We know that three vectors are co planar only if their scalar triple product is 0.

That is,

$\sf \overrightarrow {A},\overrightarrow{B},\overrightarrow{C}\:are\:coplanar\:only\:if\: \overrightarrow{A}.(\overrightarrow{B}\times \overrightarrow{C})=0$

Finding the scalar triple product of the three vectors,

$\sf \overrightarrow{A}.(\overrightarrow{B}\times \overrightarrow{C})={\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}} =a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1)$

where a₁ = 2, a₂ = 1, a₃ = 3

b₁ = -1, b₂ = -3, b₃ = -4

c₁ = 1, c₂ = -5, c₃ = -4

Substitute the data,

$\sf \overrightarrow{A}.(\overrightarrow{B}\times \overrightarrow{C})=2\times(12-20)+1\times(-4-4)+3\times(5+3)$

$\implies -16-8+24$

$\implies-24+24=0$

Hence the scalar triple product of the three vectors is 0.

Therefore the vectors are co planar.

Hence proved.