Q71 / 160 · TS EAMCET 2016 · Mathematics
The vectors $2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
- Aare linearly dependent
- Correct answerBare linearly independent
- Cform sides of a triangle
- Dare coplanar
Solution
Let $x, y$ and $z$ be the three scalars such that
$\begin{aligned}<br />& x(2 \hat{i}-3 \hat{j}+\hat{k})+y(\hat{i}-2 \hat{j}+3 \hat{k})+z(3 \hat{i}+\hat{j}-2 \hat{k})=0 \\<br />& \Rightarrow(2 x+y+3 z) \hat{i}+(-3 x-2 y+z) \hat{j}+ \\<br />& \quad(x+3 y-2 z) \hat{k}=0 \\<br />& \Rightarrow \quad 2 x+y+3 z=0,-3 x-2 y+z=0 \\<br />& \text {and } \quad x+3 y-2 z=0<br />\end{aligned}$
This is a homogeneous system of equations. The determinant of coefficient matrix is
$\left|\begin{array}{ccc}<br />2 & 1 & 3 \\<br />-3 & -2 & 1 \\<br />1 & 3 & -2<br />\end{array}\right| \neq 0$
So, the system of equations has only the trivial solution given by $x=y=z=0$.
Hence, the given vectors are linearly independent.
$\begin{aligned}<br />& x(2 \hat{i}-3 \hat{j}+\hat{k})+y(\hat{i}-2 \hat{j}+3 \hat{k})+z(3 \hat{i}+\hat{j}-2 \hat{k})=0 \\<br />& \Rightarrow(2 x+y+3 z) \hat{i}+(-3 x-2 y+z) \hat{j}+ \\<br />& \quad(x+3 y-2 z) \hat{k}=0 \\<br />& \Rightarrow \quad 2 x+y+3 z=0,-3 x-2 y+z=0 \\<br />& \text {and } \quad x+3 y-2 z=0<br />\end{aligned}$
This is a homogeneous system of equations. The determinant of coefficient matrix is
$\left|\begin{array}{ccc}<br />2 & 1 & 3 \\<br />-3 & -2 & 1 \\<br />1 & 3 & -2<br />\end{array}\right| \neq 0$
So, the system of equations has only the trivial solution given by $x=y=z=0$.
Hence, the given vectors are linearly independent.
Hence, the correct answer is option: 'B'.