Find the angle between the line \[\vec{r} = \left( 2 \hat{i}+ 3 \hat {j} + 9 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right)\] and the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 5 .\]

\[ \text{ We know that the angle θ between the line } \vec{r} = \vec{a} + \lambda \vec{b} \text{ and the plane } \vec{r} . \vec{n} =\text{ dis given by} \]\[\sin \theta = \frac{\vec{b} . \vec{n}}{\left| \vec{b} \right| \left| \vec{n} \right|} . \]\[\text{ Here} ,\]\[ \vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \text{ and } \vec{n} = \hat{i} + \hat{j} + \hat{k} \]\[ \text{ So } ,\sin \theta = \frac{\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) . \left( \hat{i} + \hat{j} + \hat{k} \right)}{\left| 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right| \left| \hat{i} + \hat{j} + \hat{k} \right|} = \frac{2 + 3 + 4}{\sqrt{4 + 9 + 16} \sqrt{1 + 1 + 1}} = \frac{9}{\sqrt{29} \sqrt{3}} = \frac{3 \sqrt{3}}{\sqrt{29}}\]\[ \Rightarrow \theta = \sin^{- 1} \left( \frac{3 \sqrt{3}}{\sqrt{29}} \right)\]

Find the Angle Between the Line → R = ( 2 ^ I + 3 ^ J + 9 ^ K ) + λ ( 2 ^ I + 3 ^ J + 4 ^ K ) and the Plane → R ⋅ ( ^ I + ^ J + ^ K ) = 5 . - Mathematics

प्रश्न

Find the angle between the line $\vec{r} = (2 \hat{i} + 3 \hat{j} + 9 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 5 .$

उत्तर

$We know that the angle θ between the line \vec{r} = \vec{a} + λ \vec{b} and the plane \vec{r} . \vec{n} = dis given by$
$\sin θ = \frac{\vec{b} . \vec{n}}{| \vec{b} | | \vec{n} |} .$
$Here,$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} and \vec{n} = \hat{i} + \hat{j} + \hat{k}$
$So, \sin θ = \frac{(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) . (\hat{i} + \hat{j} + \hat{k})}{| 2 \hat{i} + 3 \hat{j} + 4 \hat{k} | | \hat{i} + \hat{j} + \hat{k} |} = \frac{2 + 3 + 4}{\sqrt{4 + 9 + 16} \sqrt{1 + 1 + 1}} = \frac{9}{\sqrt{29} \sqrt{3}} = \frac{3 \sqrt{3}}{\sqrt{29}}$
$\Rightarrow θ = \sin^{- 1} (\frac{3 \sqrt{3}}{\sqrt{29}})$

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Angle Between Two Lines

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Q 1 Q 4 Q 2

पाठ 29: The Plane - Exercise 29.11 [पृष्ठ ६१]

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आरडी शर्मा Mathematics [English] Class 12

पाठ 29 The Plane
Exercise 29.11 | Q 1 | पृष्ठ ६१

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संबंधित प्रश्‍न

Find the acute angle between the lines whose direction ratios are 5, 12, -13 and 3, - 4, 5.

Find the angle between the following pair of lines:

$\vec{r} = 2 \hat{i} - 5 \hat{j} + \hat{k} + λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) and \vec{r} = 7 \hat{i} - 6 \hat{k} + μ (\hat{i} + 2 \hat{j} + 2 \hat{k})$

Find the angle between the following pair of lines:

$\vec{r} = 3 \hat{i} + \hat{j} - 2 \hat{k} + λ (\hat{i} - \hat{j} - 2 \hat{k}) and \vec{r} = 2 \hat{i} - \hat{j} - 56 \hat{k} + μ (3 \hat{i} - 5 \hat{j} - 4 \hat{k})$

Find the angle between the following pairs of lines:

$\frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{- 3}$ and $\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$

Find the angle between the following pairs of lines:

$\frac{x}{y} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x - 5}{4} = \frac{y - 2}{1} = \frac{z - 3}{8}$

Find the values of p so the line $\frac{1 - x}{3} = \frac{7 y - 14}{2} p = \frac{z - 3}{2}$ and $\frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ are at right angles.

Find the angle between the lines whose direction ratios are a, b, c and b − c, c − a, a − b.

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Show that the line whose vector equation is $\vec{r} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k} + λ (\hat{i} + 3 \hat{j} + 4 \hat{k})$ is parallel to the plane whose vector $\vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 7 .$ Also, find the distance between them.

Find the angle between the line $\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{2}$ and the plane

3x + 4y + z + 5 = 0.

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$\vec{r} = 2 \hat{i} - 5 \hat{j} + \hat{k} + λ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$ and $\vec{r} = 2 \hat{i} - 5 \hat{j} + \hat{k} + λ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$

$\vec{r} = 3 \hat{i} + \hat{j} + 2 \hat{k} + l (\hat{i} - \hat{j} + 2 \hat{k})$ and $\vec{r} = 2 \hat{i} + \hat{j} + 56 \hat{k} + m (3 \hat{i} - 5 \hat{j} + 4 \hat{k})$

Assertion (A): The acute angle between the line $\bar{r} = \hat{i} + \hat{j} + 2 \hat{k} + λ (\hat{i} - \hat{j})$ and the x-axis is $\frac{π}{4}$

Reason(R): The acute angle 𝜃 between the lines $\bar{r} = x_{1} \hat{i} + y_{1} \hat{j} + z_{1} \hat{k} + λ (a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k})$ and $\bar{r} = x_{2} \hat{i} + y_{2} \hat{j} + z_{2} \hat{k} + μ (a_{2} \hat{i} + b_{2} \hat{j} + c_{2} \hat{k})$ is given by cosθ = $\frac{| a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} |}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}}$

The angle between two lines $\frac{x + 1}{2} = \frac{y + 3}{2} = \frac{z - 4}{- 1}$ and $\frac{x - 4}{1} = \frac{y + 4}{2} = \frac{z + 1}{2}$ is ______.

Find the angle between the following two lines:

$\vec{r} = 2 \hat{i} - 5 \hat{j} + \hat{k} + λ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$

$\vec{r} = 7 \hat{i} - 6 \hat{k} + μ (\hat{i} + 2 \hat{j} + 2 \hat{k})$

Find the Angle Between the Line → R = ( 2 ^ I + 3 ^ J + 9 ^ K ) + λ ( 2 ^ I + 3 ^ J + 4 ^ K ) and the Plane → R ⋅ ( ^ I + ^ J + ^ K ) = 5 . - Mathematics

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Find the Angle Between the Line → R = ( 2 ^ I + 3 ^ J + 9 ^ K ) + λ ( 2 ^ I + 3 ^ J + 4 ^ K ) and the Plane → R ⋅ ( ^ I + ^ J + ^ K ) = 5 . - Mathematics

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