The position vector of the point of intersect | vedclass

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(D) The equation of the line is given by $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$.Any point on this line can be re

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$4 \hat{i}-3 \hat{j}$

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(D) The equation of the line is given by $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$.Any point on this line can be represented in Cartesian coordinates as $(x, y, z) = (2+\lambda, 1-2\lambda, -4+2\lambda)$.The $XOY$-plane is defined by the equation $z=0$.Since the point of intersection lies on the $XOY$-plane,we set the $z$-coordinate to zero:$-4+2\lambda = 0 \Rightarrow 2\lambda = 4 \Rightarrow \lambda = 2$.Substituting $\lambda = 2$ back into the coordinates:$x = 2+2 = 4$$y = 1-2(2) = 1-4 = -3$$z = -4+2(2) = 0$.Thus,the point of intersection is $(4, -3, 0)$,and its position vector is $4 \hat{i}-3 \hat{j}$.

The position vector of the point of intersection of the line $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and the $XOY$-plane is:

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