Find the vector equation of the plane passing through the intersection of the planes
\(\overrightarrow r.(2\hat i+2\hat j-3\hat k)=7\),\(\overrightarrow r.(2\hat i+5\hat j+3\hat k)=9\) and through the point ( 2, 1, 3 ).
The equations of the planes are \(\overrightarrow r.(2\hat i+2\hat j-3\hat k)=7\) and \(\overrightarrow r.(2\hat i+5\hat j+3\hat k)=9\)
\(\Rightarrow \overrightarrow r.(2\hat i+2\hat j-3\hat k)-7=0\)...(1)
\(\Rightarrow\overrightarrow r.(2\hat i+5\hat j+3\hat k)-9=0\)...(2)
The equation of any plane through the intersection of the planes given in equation (1) and (2) is given by,
[ \(\overrightarrow r.(2\hat i+2\hat j-3\hat k)=7\) ]+\(\lambda\) [ \(\overrightarrow r.(2\hat i+5\hat j+3\hat k)=9\) ]=0, where \(\lambda\in\) R
\(\overrightarrow r.[(2\hat i+2\hat j-3\hat k)\)+\(\lambda(2\hat i+5\hat j+3\hat k)]=9\lambda+7\)
\(\overrightarrow r[(2+2\lambda)\hat i+(2+5\lambda)\hat j+(3\lambda-3)\hat k]=9\lambda+7\) ...(3)
The plane passes through the point (2,1, 3).
Therefore, its position vector is given by, \(\overrightarrow r.2\hat i+2\hat j+3\hat k\)
Substituting in equation (3), we obtain
\((2\hat i+2\hat j-3\hat k).\)\([(2+2\lambda)\hat i+(2+5\lambda)\hat j+(3\lambda-3)\hat k]=9\lambda+7\)
\(\Rightarrow[(2+2\lambda)+(2+5\lambda)+(3\lambda-3)]=9\lambda+7\)
\(\Rightarrow\) 18λ-3=9λ+7
\(\Rightarrow\) 9λ=10
\(\Rightarrow\) λ=\(\frac{10}{9}\)
Substituting \(\lambda=\frac{10}{9}\) in equation(3), we obtain
\(\overrightarrow r\bigg(\frac{38}{9}\hat i+\frac{68}{9}\hat j+\frac{3}{9}\hat k\bigg)=17\)
\(\Rightarrow \overrightarrow r (38\hat i+68\hat j+3\hat k)=153\)
This is the vector equation of the required plane.
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely: