Question

The values of $\lambda$ such that $(x,y,z)\ne (0,0,0)$ and $(\hat{i}+\hat{j}+3\hat{k})x+(3\hat{i}-3\hat{j}+\hat{k})y+(-4\hat{i}+5\hat{j})z$
$=\lambda \left(\hat{i}x+\hat{j}y+\hat{k}z\right)$ are

• A. 0, 1
• B. -1, 1
• C. -1, 0
• D. -2, 0

Answer 1

Answer: (C)

Given,
$(\hat{i}+\hat{j}+3\hat{k}{)}_{x+}(3\hat{i}-3\hat{j}+\hat{k}{)}_{y+}(-4\hat{i}+5\hat{j}{)}_z=\lambda (\hat{i}x+\hat{j}y+\hat{k}z)$
On equating the coefficients of $\hat{i},\hat{j}$ and $k$ both sides, we have
$x+3y-4z=\lambda x$
$x-3y+5z=\lambda y$
and $3x+y+0=\lambda z$
Above three equations can be rewritten as
$(1-\lambda )x+3y-4z=0$
$x-(3+\lambda )y+5z=0$
$3x+y-\lambda z=0$
This is homogeneous system of equations in three variables $x,y$ and $z$.
It is consistent and have non-zero solution.
i.e. $(x,y,z)\ne (0,0,0)$, if determinant of coefficient matrix is zero.
$\Rightarrow \left|\begin{array}{ccc}1-\lambda & 3& -4\\ 1& -(3+\lambda )& 5\\ 3& 1& -\lambda \end{array}\right|=0$
On expanding along first row, we have
$(1-\lambda )[\lambda (3+\lambda )-5]-3(-\lambda -15)-4(1+9+3\lambda )=0$
$\Rightarrow (1-\lambda )\left({\lambda }^2+3\lambda -5\right)+3\lambda +45-40-12\lambda =0$
$\Rightarrow {\lambda }^2+3\lambda -5-{\lambda }^3-3{\lambda }^2+5\lambda -9\lambda +5=0$
$\Rightarrow -{\lambda }^3-2{\lambda }^2-\lambda =0$
$\Rightarrow \lambda \left({\lambda }^2+2\lambda +1\right)=0$
$\Rightarrow \lambda (\lambda +1{)}^2=0$
$\Rightarrow \lambda =0,-1$