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Vector Algebra Part 3

Section: Mathematics

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Question : 65 of 72

Marks: +1, -0

The distance of the point

Q(0,2,-2)

form the line passing through the point

P(5,-4,3)

and perpendicular to the lines

\vec{r}=(-3\hat{i}+2\hat{k})+

\lambda(2\hat{i}+3\hat{j}+5\hat{k}), \;\; \lambda\in\mathbb{R} \;\;

\;\; \vec{r}=(\hat{i}-2\hat{j}+\hat{k})+

\mu(-\hat{i}+3\hat{j}+2\hat{k}), \mu\in\mathbb{R}

[31-Jan-2024 Shift 1]

$\sqrt{86}$
$\sqrt{20}$
$\sqrt{54}$
$\sqrt{74}$

Solution: 👈: Video Solution

A vector in the direction of the required line can be obtained by cross product of

\;\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix}

\;=-9\hat{i}-9\hat{j}+9\hat{k}

Required line,

\;\vec{r}=(5\hat{i}-4\hat{j}+3\hat{k})+\lambda'(-9\hat{i}-9\hat{j}+9\hat{k})

\;\vec{r}=(5\hat{i}-4\hat{j}+3\hat{k})+\lambda(\hat{i}+\hat{j}-\hat{k})

Now distance of

(0,2,-2)

\;P.V. \;\text{ of }\; P\equiv(5+\lambda)\hat{i}+(\lambda-4)\hat{j}

+(3-\lambda)\hat{k}

\;\vec{AP}=(5+\lambda)\hat{i}+(\lambda-6)\hat{j}+(5-\lambda)\hat{k}

\;\vec{AP}\cdot(\hat{i}+\hat{j}-\hat{k})=0

\;5+\lambda+\lambda-6-5+\lambda=0

\;\lambda=2

\;|\vec{AP}|=\sqrt{49+16+9}

\;|\vec{AP}|=\sqrt{74}

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