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Question

For the following problem, find the point where the given line intersects the given plane.

Line: r=(4+13d)t+(1+23d)j+(7+23d)k.\vec{r}=\left(4+\frac{1}{3} d\right) \vec{t}+\left(1+\frac{2}{3} d\right)\vec {j}+\left(7+\frac{2}{3} d\right) \vec{k}.

Plane: x+4y3z=35x+4 y-3 z=35

Solution

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r=(4+13d)i^+(1+23d)j^+(7+23d)k^\overline{r}=(4+\dfrac{1}{3}d)\widehat{i}+(1+\dfrac{2}{3}d)\widehat{j}+(7+\dfrac{2}{3}d)\widehat{k}

x+4y3z=35x+4y-3z=35

Given\text{\textcolor{#c34632}{Given}}

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