The ratio in which the plane \overrightarrow{r} \cdot \left (\overrightarrow{i}  - 2 \overrightarrow{j}  + 3 \overrightarrow{k}   \right ) = 17 divides the line joining the points -2 \overrightarrow{i}  + 4 \overrightarrow{j}  + 7 \overrightarrow{k}  and 3 \overrightarrow{i}  - 5 \overrightarrow{j}  + 8 \overrightarrow{k} 

Engineering

Mathematics

Line of Greatest Slope

Question

The ratio in which the plane

\overrightarrow{r} \cdot&nbsp;\left (\overrightarrow{i}&nbsp; - 2 \overrightarrow{j}&nbsp; + 3 \overrightarrow{k}&nbsp;&nbsp; \right ) = 17

divides the line joining the points

-2 \overrightarrow{i}&nbsp; + 4 \overrightarrow{j}&nbsp; + 7 \overrightarrow{k}&nbsp;

and

3&nbsp;\overrightarrow{i}&nbsp; - 5 \overrightarrow{j}&nbsp; + 8 \overrightarrow{k}&nbsp;

1 : 5

1 : 10

3 : 5

3 : 10

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Let the plane

\vec{r} \cdot (\vec{i}, - 2 \vec{j}, + 3 \vec{k}) = 17

divided the line joining the points

- 2 \vec{i}, + 4 \vec{j}, + 7 \vec{k}

and

3 \vec{i}, - 5 \vec{j}, + 8 \vec{k}

in the ratio

t : 1

at point P.

Therefore, point P is

\frac{3 t - 2}{t + 1} \vec{i} +, \frac{- 5 t + 4}{t + 1} \vec{j} + \frac{8 t + 7}{t + 1} \vec{k}

This lies on the given plane

∴ \frac{3 t - 2}{t + 1} (1) +, \frac{- 5 t + 4}{t + 1} (- 2) + \frac{8 t + 7}{t + 1} (3) = 17

Solving, we get

t = \frac{3}{10}

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