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Ncert Solutions Maths class 12th Maths Class 12th Three Dimensional Geometry

55. Find the vector equation of the line passing through (1, 2, 3) and parallel to the plane $\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5$ and $\vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6$

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Vishal Baghel | Contributor-Level 10

4 months ago

Let the required line be parallel to vector $\vec{b}$ given by,

$\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, 3) is $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

The equation of line passing through (1, 2, 3) and parallel to $\vec{b}$ is given by,

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . . (1) \end{array}$

The equations of the given planes are

$\begin{array}{l} \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5 . . . . . . . . (2) \\ \vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6 . . . . . . . . (3) \end{array}$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

$\begin{array}{l} \Rightarrow (\hat{i} - \hat{j} + 2 \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0 \\ \Rightarrow λ (b_{1} - b_{2} + 2 b_{3}) = 0 \\ \Rightarrow b_{1} - b_{2} + 2 b_{3} = 0 . . . . . . . . . . (4) \end{array}$

Similarly, $(3 \hat{i} + \hat{j} + \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 1) \times 1 - 1 \times 2} = \frac{b_{2}}{2 \times 3 - 1 \times 1} = \frac{b_{3}}{1 \times 1 - 3 \times (- 1)} \\ \Rightarrow \frac{b_{1}}{- 3} = \frac{b_{2}}{5} = \frac{b_{3}}{4} \end{array}$

Therefore, the direction ratios of $\vec{b}$ are −3, 5, and 4.

$∴ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = - 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$

Substituting the value of $\vec{b}$ in equation (1), we obtain

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 3 \hat{i} + 5 \hat{j} + 4 \hat{k})$

This is the e

...more

Let the required line be parallel to vector $\vec{b}$ given by,

$\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, 3) is $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

The equation of line passing through (1, 2, 3) and parallel to $\vec{b}$ is given by,

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . . (1) \end{array}$

The equations of the given planes are

$\begin{array}{l} \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5 . . . . . . . . (2) \\ \vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6 . . . . . . . . (3) \end{array}$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

Similarly, $(3 \hat{i} + \hat{j} + \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0$

From equations (4) and (5), we obtain

Therefore, the direction ratios of $\vec{b}$ are −3, 5, and 4.

$∴ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = - 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$

Substituting the value of $\vec{b}$ in equation (1), we obtain

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 3 \hat{i} + 5 \hat{j} + 4 \hat{k})$

This is the equation of the required line.

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If $\int_{}^{} (\frac{x^{2} c o s x}{1 + π^{x}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) d x = \frac{π}{4} (π + α) - 2$

Then the value of 'α' is equal to

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$\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{x^{2} c o s x}{1 + π^{2}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) d x = \frac{π}{4} (π + α) - 2$

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Given $| \vec{a} | = 1$ , $| \vec{b} | = 4$ , $\vec{a} \cdot \vec{b} = 2$

$\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$

Dot product with $\vec{a}$ on both sides

$\vec{c} \cdot \vec{a} = - 6$ ... (1)

Dot product with $\vec{b}$ on both sides

$\vec{b} \cdot \vec{c} = - 48$ ... (2)

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If the foot of perpendicular from (1, 2, 3) to the line $\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z - 4}{1}$ is (a, b, g) then find a + b + g

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(a – 1) × 2 + (b – 2) × 5 + (g – 3) × 1 = 0

2a + 5b + g – 15 = 0

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a + 1 = 2λ

b – 2 = 5λ

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55. Find the vector equation of the line passing through (1, 2, 3) and parallel to the plane $\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5$ and $\vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6$

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55. Find the vector equation of the line passing through (1, 2, 3) and parallel to the plane r→=(i^−j^+2k^)=5 and r→.(3i^+j^+k^)=6

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55. Find the vector equation of the line passing through (1, 2, 3) and parallel to the plane $\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5$ and $\vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6$