A line L is passing through the point A whose | vedclass

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(A) The vector equation of line $L$ passing through $A(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k}$ is $\vec{r}

Vedclass · Accepted Answer

$\frac{1}{3}(-7 \hat{i}+\hat{j}-19 \hat{k})$

Vedclass · Answer

(A) The vector equation of line $L$ passing through $A(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k}$ is $\vec{r} = (1+2\lambda)\hat{i} + (2+\lambda)\hat{j} + (-3+2\lambda)\hat{k}$.The plane $\pi$ passes through $P_1(\hat{i}+\hat{j}+\hat{k})$ and $P_2(\hat{i}-\hat{j}-\hat{k})$ and is parallel to $\vec{v}=\hat{i}-2\hat{j}$.The normal vector to the plane is $\vec{n} = (P_2 - P_1) 	imes \vec{v} = (0\hat{i}-2\hat{j}-2\hat{k}) 	imes (1\hat{i}-2\hat{j}+0\hat{k}) = -4\hat{i} - 2\hat{j} + 2\hat{k}$.We can use the simplified normal vector $\vec{n}' = 2\hat{i} + \hat{j} - \hat{k}$.The equation of the plane is $2(x-1) + 1(y-1) - 1(z-1) = 0 \Rightarrow 2x + y - z = 2$.Substituting the coordinates of the line $L$ into the plane equation:$2(1+2\lambda) + (2+\lambda) - (-3+2\lambda) = 2$$2 + 4\lambda + 2 + \lambda + 3 - 2\lambda = 2$$3\lambda + 7 = 2 \Rightarrow 3\lambda = -5 \Rightarrow \lambda = -\frac{5}{3}$.Substituting $\lambda = -\frac{5}{3}$ into the line equation:$x = 1 + 2(-\frac{5}{3}) = -\frac{7}{3}$, $y = 2 - \frac{5}{3} = \frac{1}{3}$, $z = -3 + 2(-\frac{5}{3}) = -\frac{19}{3}$.Thus, the point is $\frac{1}{3}(-7\hat{i} + \hat{j} - 19\hat{k})$.

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