The distance of the point Q(0, 2, –2) from the line passing through the point P(5, –4, 3) and perpendicular to the lines and is : [2024]

Question

The distance of the point Q(0, 2, –2) from the line passing through the point P(5, –4, 3) and perpendicular to the lines $\vec{r} = (- 3 \hat{i} + 2 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 5 \hat{k}), λ \in R$ and $\vec{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + μ (- \hat{i} + 3 \hat{j} + 2 \hat{k}), μ \in R$ is : [2024]

Smart Achievers · Accepted Answer

(1)

Given, $\vec{r} = (- 3 \hat{i} + 2 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 5 \hat{k})$

and $\vec{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + μ (- \hat{i} + 3 \hat{j} + 2 \hat{k})$

${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ - 1 & 3 & 2 \end{matrix} |$

${\vec{b}}_{1} \times {\vec{b}}_{2} = \hat{i} (6 - 15) - \hat{j} (4 + 5) + \hat{k} (6 + 3)$

${\vec{b}}_{1} \times {\vec{b}}_{2} = - 9 \hat{i} - 9 \hat{j} + 9 \hat{k}$

Then the equation of the line

$l = \frac{x - 5}{1} = \frac{y + 4}{1} = \frac{z - 3}{- 1} = λ$

$x = λ + 5; y = λ - 4; z = 3 - λ$

$∴ \vec{Q R} \cdot l = 0$

$(λ + 5) \cdot 1 + (λ - 6) 1 + (3 - λ + 2) (- 1) = 0$

$λ + 5 + λ - 6 + (- 5) + λ = 0$

$λ + 5 + λ - 6 - 5 + λ = 0$

$3 λ - 6 = 0$

$λ = \frac{6}{3}$

$λ = 2$

Then, R = (7, –2, 1)

$Q R = \sqrt{49 + 16 + 9} \Rightarrow Q R = \sqrt{74}$ .

Solution

1 $\sqrt{74}$

2 $\sqrt{54}$

3 $\sqrt{86}$

4 $\sqrt{20}$

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Solution

Q. The distance of the point Q(0, 2, –2) from the line passing through the point P(5, –4, 3) and perpendicular to the lines r→=(–3i^+2k^)+λ(2i^+3j^+5k^), λ∈R and r→=(i^–2j^+k^)+μ(–i^+3j^+2k^), μ∈R is : [2024]

1 74

2 54

3 86

4 20

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1 $\sqrt{74}$

2 $\sqrt{54}$

3 $\sqrt{86}$

4 $\sqrt{20}$