In what ratio does the plane vec{r} cdot (hat | vedclass

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Question

(C) Let the required ratio be $\lambda : 1$. The position vector of the point dividing the line segment is given by the section formula:$\vec{r} = \fr

Vedclass · Accepted Answer

$3 : 10$

Vedclass · Answer

(C) Let the required ratio be $\lambda : 1$. The position vector of the point dividing the line segment is given by the section formula:$\vec{r} = \frac{(\lambda)(3\hat{i} - 5\hat{j} + 8\hat{k}) + (1)(-2\hat{i} + 4\hat{j} + 7\hat{k})}{\lambda + 1}$$= \left( \frac{3\lambda - 2}{\lambda + 1} \right)\hat{i} + \left( \frac{-5\lambda + 4}{\lambda + 1} \right)\hat{j} + \left( \frac{8\lambda + 7}{\lambda + 1} \right)\hat{k}$Since this point lies on the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 17$,we substitute the coordinates into the plane equation:$\left( \frac{3\lambda - 2}{\lambda + 1} \right)(1) + \left( \frac{-5\lambda + 4}{\lambda + 1} \right)(-2) + \left( \frac{8\lambda + 7}{\lambda + 1} \right)(3) = 17$$(3\lambda - 2) + 10\lambda - 8 + 24\lambda + 21 = 17(\lambda + 1)$$37\lambda + 11 = 17\lambda + 17$$20\lambda = 6$$\lambda = \frac{6}{20} = \frac{3}{10}$Thus,the required ratio is $3 : 10$.

In what ratio does the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 17$ divide the line segment joining the points $-2\hat{i} + 4\hat{j} + 7\hat{k}$ and $3\hat{i} - 5\hat{j} + 8\hat{k}$?

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