The distance from the point -i + 2j + 6k to t | vedclass

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(A) Let $A$ be the point $(-1, 2, 6)$ and $P$ be the point $(2, 3, -4)$ on the line. The vector $\overrightarrow{PA}$ is given by $\overrightarrow{A}

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let $A$ be the point $(-1, 2, 6)$ and $P$ be the point $(2, 3, -4)$ on the line. The vector $\overrightarrow{PA}$ is given by $\overrightarrow{A} - \overrightarrow{P} = (-1-2)i + (2-3)j + (6-(-4))k = -3i - j + 10k$.The line is parallel to the vector $\vec{v} = 6i + 3j - 4k$. The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.The projection of $\overrightarrow{PA}$ on the line is given by $PN = \left| \frac{\overrightarrow{PA} \cdot \vec{v}}{|\vec{v}|} ight| = \left| \frac{(-3)(6) + (-1)(3) + (10)(-4)}{\sqrt{61}} ight| = \left| \frac{-18 - 3 - 40}{\sqrt{61}} ight| = \left| \frac{-61}{\sqrt{61}} ight| = \sqrt{61}$.In the right-angled triangle $	riangle APN$,the distance $AN$ from point $A$ to the line is given by $AN = \sqrt{|\overrightarrow{PA}|^2 - PN^2}$.We have $|\overrightarrow{PA}|^2 = (-3)^2 + (-1)^2 + 10^2 = 9 + 1 + 100 = 110$.Thus,$AN = \sqrt{110 - 61} = \sqrt{49} = 7$.

The distance from the point $-i + 2j + 6k$ to the straight line passing through the point $(2, 3, -4)$ and parallel to the vector $6i + 3j - 4k$ is

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