Find the distance of the point (-1, -5, -10) | vedclass

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(A) The equation of the line is $\vec{r} = (2 + 3\lambda)\hat{i} + (-1 + 4\lambda)\hat{j} + (2 + 2\lambda)\hat{k}$.Substituting this into the plane eq

Vedclass · Accepted Answer

$13$

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(A) The equation of the line is $\vec{r} = (2 + 3\lambda)\hat{i} + (-1 + 4\lambda)\hat{j} + (2 + 2\lambda)\hat{k}$.Substituting this into the plane equation $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$:$((2 + 3\lambda)\hat{i} + (-1 + 4\lambda)\hat{j} + (2 + 2\lambda)\hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$$(2 + 3\lambda) - (-1 + 4\lambda) + (2 + 2\lambda) = 5$$2 + 3\lambda + 1 - 4\lambda + 2 + 2\lambda = 5$$5 + \lambda = 5 \implies \lambda = 0$.Substituting $\lambda = 0$ into the line equation,the point of intersection is $(2, -1, 2)$.The distance between $(2, -1, 2)$ and $(-1, -5, -10)$ is given by the distance formula:$d = \sqrt{(-1 - 2)^2 + (-5 - (-1))^2 + (-10 - 2)^2}$$d = \sqrt{(-3)^2 + (-4)^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

Find the distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$.

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