The distance of the point ({x_1}, {y_1}, {z_1 | vedclass

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Question

(A) Let $P = ({x_1}, {y_1}, {z_1})$ be the point and $A = ({x_2}, {y_2}, {z_2})$ be a point on the line. The vector $\vec{AP} = ({x_1} - {x_2})\hat{i}

Vedclass · Accepted Answer

$\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2} - {{[l({x_1} - {x_2}) + m({y_1} - {y_2}) + n({z_1} - {z_2})]}^2}}$

Vedclass · Answer

(A) Let $P = ({x_1}, {y_1}, {z_1})$ be the point and $A = ({x_2}, {y_2}, {z_2})$ be a point on the line. The vector $\vec{AP} = ({x_1} - {x_2})\hat{i} + ({y_1} - {y_2})\hat{j} + ({z_1} - {z_2})\hat{k}$.Let $\vec{u} = l\hat{i} + m\hat{j} + n\hat{k}$ be the unit vector along the line.The projection of $\vec{AP}$ on the line is $p = |\vec{AP} \cdot \vec{u}| = |l({x_1} - {x_2}) + m({y_1} - {y_2}) + n({z_1} - {z_2})|$.By the Pythagorean theorem,the perpendicular distance $d$ is given by $d^2 = |\vec{AP}|^2 - p^2$.$|\vec{AP}|^2 = ({x_1} - {x_2})^2 + ({y_1} - {y_2})^2 + ({z_1} - {z_2})^2$.Thus,$d = \sqrt{({x_1} - {x_2})^2 + ({y_1} - {y_2})^2 + ({z_1} - {z_2})^2 - [l({x_1} - {x_2}) + m({y_1} - {y_2}) + n({z_1} - {z_2})]^2}$.

The distance of the point $({x_1}, {y_1}, {z_1})$ from the line $\frac{{x - {x_2}}}{l} = \frac{{y - {y_2}}}{m} = \frac{{z - {z_2}}}{n}$,where $l, m, n$ are the direction cosines of the line,is given by:

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