The projection of the line segment joining th | vedclass

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(B) Let the points be $A(-1, 0, 3)$ and $B(2, 5, 1)$.The vector $\vec{AB} = (2 - (-1))\hat{i} + (5 - 0)\hat{j} + (1 - 3)\hat{k} = 3\hat{i} + 5\hat{j}

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$\frac{22}{7}$

Vedclass · Answer

(B) Let the points be $A(-1, 0, 3)$ and $B(2, 5, 1)$.The vector $\vec{AB} = (2 - (-1))\hat{i} + (5 - 0)\hat{j} + (1 - 3)\hat{k} = 3\hat{i} + 5\hat{j} - 2\hat{k}$.The direction ratios of the line are $6, 2, 3$. The magnitude of this direction vector is $\sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.The direction cosines of the line are $l = \frac{6}{7}, m = \frac{2}{7}, n = \frac{3}{7}$.The projection of the line segment $AB$ on the line is given by the dot product of the vector $\vec{AB}$ and the unit vector along the line:Projection $= (x_2 - x_1)l + (y_2 - y_1)m + (z_2 - z_1)n$$= (3)(\frac{6}{7}) + (5)(\frac{2}{7}) + (-2)(\frac{3}{7})$$= \frac{18 + 10 - 6}{7} = \frac{22}{7}$.

The projection of the line segment joining the points $(-1, 0, 3)$ and $(2, 5, 1)$ on the line whose direction ratios are $6, 2, 3$ is

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