Question

The cartesian equation of the plane whose vector equation is
$\gamma=(1+\lambda-\mu) \hat{ i }+(2-\lambda) \hat{ j }+(3-2 \lambda+2 \mu) \hat{ k }$
where $\lambda, \mu$ are scalars, is

• A. 2 x+y=5
• B. 2 x-y=5
• C. 2 x-z=5
• D. 2 x+z=5

Answer 1

Answer: (D)

Given,
$\gamma =(1+\lambda-\mu) \hat{ i }+(2-\lambda) \hat{ j }+(3-2 \lambda+2 \mu) \hat{ k }$
$=(\hat{ i }+2 \hat{ j }+3 \hat{ k })+\lambda(\hat{ i }-\hat{ j }-2 \hat{ k })+\mu(-\hat{ i }+2 \hat{ k })$
Equation of plane is given by
$ \{ r -(\hat{ i }+2 \hat{ j }+3 \hat{ k })\} \cdot[(\hat{ i }-\hat{ j }-2 \hat{ k }) \times(-\hat{ i }+2 \hat{ k })]=0 $
$\Rightarrow {[ r -(\hat{ i }+2 \hat{ j }+3 \hat{ k })] \cdot(-2 \hat{ i }-\hat{ k })=0} $
$\Rightarrow r \cdot(2 \hat{ i }+\hat{ k })-5 =0$
$\Rightarrow r \cdot(2 \hat{ i }+\hat{ k })=5$
Now, equation of plane in cartesian form is
$ (x \hat{ i }+y \hat{ j }+z \hat{ k }) \cdot(2 \hat{ i }+\hat{ k })=5$
$\therefore 2 x+z=5 $