Vectors 5 Question 16
16. Find all values of $\lambda$, such that $x, y, z \neq(0,0,0)$ and $(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) x+(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) y$ $+(-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) z=\lambda(\hat{\mathbf{i}} x+\hat{\mathbf{j}} y+\hat{\mathbf{k}} z)$, where $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$ are unit vectors along the coordinate axes.
(1982, 2M)
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Solution:
- Since, $(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) x+(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) y+(-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) z$
$$ =\lambda(\hat{\mathbf{i}} x+\hat{\mathbf{j}} y+\hat{\mathbf{k}} z) $$
$\Rightarrow x+3 y-4 z=\lambda x, x-3 y+5 z=\lambda y, 3 x+y+0 z=\lambda z$
$\Rightarrow(1-\lambda) x+3 y-4 z=0, x-(3+\lambda) y+5 z=0$,
$$ 3 x+y-\lambda z=0 $$
Since, $\quad(x, y, z) \neq(0,0,0)$
$\therefore$ Non-trivial solution.
$$ \begin{aligned} & \Rightarrow \quad \Delta=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} 1-\lambda & 3 & -4 \\ 1 & -(3+\lambda) & 5 \\ 3 & 1 & -\lambda \end{array}\right|=0 \\ & \Rightarrow(1-\lambda)\left(3 \lambda+\lambda^{2}-5\right)-3(-\lambda-15)-4(1+9+3 \lambda)=0 \\ & \Rightarrow-\lambda^{3}-2 \lambda^{2}-\lambda=0 \Rightarrow \lambda(\lambda+1)^{2}=0 \\ & \therefore \quad \lambda=0,-1 \end{aligned} $$
