Question

The vector equation of the plane passing through the intersection of the planes $\vec{ r } .(\hat{ i }+\hat{ j }+\hat{ k })=1$ and $\vec{ r } .(\hat{ i }-2 \hat{ j })=-2,$ and the point (1,0,2) is :

• A. $\vec{ r } .(\hat{ i }+7 \hat{ j }+3 \hat{ k })=\frac{7}{3}$
• B. $\vec{ r } \cdot(3 \hat{ i }+7 \hat{ j }+3 \hat{ k })=7$
• C. $\vec{ r } \cdot(\hat{ i }+7 \hat{ j }+3 \hat{ k })=7$
• D. $\vec{ r } .(\hat{ i }-7 \hat{ j }+3 \hat{ k })=\frac{7}{3}$

Answer 1

The Correct Option is C

$\vec{ r } \cdot(\hat{ i }+\hat{ j }+\hat{ k })=1$ $\vec{ r } \cdot(\hat{ i }-2 \hat{ j })=-2$ point (1,0,2) Eq $^{n}$ of plane $\vec{ r } \cdot(\hat{ i }+\hat{ j }+\hat{ k })-1+\lambda\{ r .(\hat{ i }-2 \hat{ j })+2\}=0$ $\vec{ r } \cdot\{\hat{ i }(1+\lambda)+\hat{ j }(1-2 \lambda)+\hat{ k }(1)\}-1+2 \lambda=0$ Point $\hat{ i }+0 \hat{ j }+2 \hat{ k }=\vec{ r }$ $\therefore(\hat{ i }+2 \hat{ k }) \cdot\{\hat{ i }(1+\lambda)+\hat{ j }(1-2 \lambda)+\hat{ k }(1)\} -1+2 \lambda=0$ $1+\lambda+2-1+2 \lambda=0$ $\lambda=-\frac{2}{3}$ $\therefore\,\,\,\, \vec{ r } \cdot\left[\hat{ i }\left(\frac{1}{3}\right)+\hat{ j }\left(\frac{7}{3}\right)+\hat{ k}\right]$ $=\frac{7}{3}$ $r\cdot[\hat{ i }+7 \hat{ j }+3 \hat{ k }]=7$