The position vector of a point P is 2 hat{i}+ | vedclass

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(A) Given, the position vector of point $P$ is $\vec{p} = 2 \hat{i}+\hat{j}+3 \hat{k}$.The plane $\pi$ is determined by vectors $\vec{a} = -\hat{i}-2

Vedclass · Accepted Answer

$r=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(-\hat{i}+5 \hat{j}-2 \hat{k})$

Vedclass · Answer

(A) Given, the position vector of point $P$ is $\vec{p} = 2 \hat{i}+\hat{j}+3 \hat{k}$.The plane $\pi$ is determined by vectors $\vec{a} = -\hat{i}-2 \hat{k}$ and $\vec{b} = \hat{i}+\hat{j}+2 \hat{k}$.The normal vector to the plane is $\vec{n} = \vec{a} 	imes \vec{b}$.The line lies on the plane and is normal to $\vec{b}$, so its direction vector $\vec{v}$ must be perpendicular to both $\vec{n}$ and $\vec{b}$.Thus, $\vec{v} = \vec{n} 	imes \vec{b} = (\vec{a} 	imes \vec{b}) 	imes \vec{b}$.Using the vector triple product formula $(\vec{a} 	imes \vec{b}) 	imes \vec{b} = (\vec{a} \cdot \vec{b})\vec{b} - (\vec{b} \cdot \vec{b})\vec{a}$.Calculate $\vec{a} \cdot \vec{b} = (-1)(1) + (0)(1) + (-2)(2) = -1 - 4 = -5$.Calculate $\vec{b} \cdot \vec{b} = (1)^2 + (1)^2 + (2)^2 = 1 + 1 + 4 = 6$.So, $\vec{v} = -5(\hat{i}+\hat{j}+2 \hat{k}) - 6(-\hat{i}-2 \hat{k}) = -5\hat{i}-5\hat{j}-10\hat{k} + 6\hat{i} + 12\hat{k} = \hat{i}-5\hat{j}+2\hat{k}$.Note: The direction vector can be scaled by $-1$, giving $-\hat{i}+5\hat{j}-2\hat{k}$.The equation of the line is $\vec{r} = \vec{p} + \lambda \vec{v} = 2 \hat{i}+\hat{j}+3 \hat{k} + \lambda(-\hat{i}+5 \hat{j}-2 \hat{k})$.

The position vector of a point $P$ is $2 \hat{i}+\hat{j}+3 \hat{k}$ and $a=-\hat{i}-2 \hat{k}, b=\hat{i}+\hat{j}+2 \hat{k}$ are two vectors which determine a plane $\pi$. The equation of a line through $P$ normal to $b$ and lying on the plane $\pi$ is

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