24. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector <math><mrow><mn>3</mn><mover accent="true"><mrow><mi>i</mi></mrow><mo stretchy="true">^</mo></mover><mo>+</mo><mn>5</mn><mover accent="true"><mrow><mi>j</mi></mrow><mo stretchy="true">^</mo></mover><mo>−</mo><mn>6</mn><mover accent="true"><mrow><mi>k</mi></mrow><mo stretchy="true">^</mo></mover></mrow></math>

(a) It is given that equation of the plane isr→.(i^+j^−k^)=2For any arbitrary point P(x, y, z) on the plane, position vector r→ I s given by, r→=xi^+yj^−zk^Substituting the value of r→ in equation (1), we obtain(xi^+yj^−zk^).(i^+j^−k^)=2⇒x+y−z=2This is the Cartesian equation of the plane.(b) r→.(2i^+3j^−4k^)=1For any arbitrary point P(x, y, z) on the plane, position vector r→ is given by, r→=(xi^+yj^−zk^)Substituting the value of r→ in equation (1), we obtain(xi^+yj^−zk^).(2i^+3j^−4k^)=1⇒2x+3y−4z=1This is the Cartesian equation of the plane.(c) r→.[(s−2t)i^+(3−t)j^+(2s+t)k^]=15For any arbitrary point P(x, y, z) on the plane, position vector r→ is given by, r→=(xi^+yj^−zk^)Substituting the value of r→ in equation (1), we obtain(xi^+yj^−zk^).[(s−2t)i^+(3−t)j^+(2s+t)k^]=15⇒(s−2t)x+(3−t)y+(2s+t)z=15This is the Cartesian equation of the given plane.

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

24. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$

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Vishal Baghel | Contributor-Level 10

4 months ago

(a) It is given that equation of the plane is

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

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value

...more

(a) It is given that equation of the plane is

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

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

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

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

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

$\begin{array}{l} (x \hat{i} + y \hat{j} - z \hat{k}) . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \\ \Rightarrow (s - 2 t) x + (3 - t) y + (2 s + t) z = 15 \end{array}$

This is the Cartesian equation of the given plane.

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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

alok kumar singh

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

$\int_{0}^{\frac{π}{2}} {(\frac{x^{2} c o s x}{1 + π^{x}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) + (\frac{x^{2} c o s x}{1 + π^{- x}} + \frac{1 + s i n^{2} x}{1 + e^{- {(s i n x)}^{2023}}})} d x$

$= \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} (x^{2} c o s x + 1 + s i n^{2} x) d x = \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} x^{2} c o s x d x + \int_{0}^{\frac{π}{2}} (1 + s i n^{2} x) d x = \frac{π}{4} (π + α) - 2$ ....(1)

Let $I_{1} = \int_{0}^{\frac{π}{2}} (1 + s i n^{2} x) d x$

$I_{1} = \int_{0}^{\frac{π}{2}} 1 \cdot d x + \int_{0}^{\frac{π}{2}} (\frac{1 - c o s 2 x}{2}) d x$

$I_{1} = \frac{π}{2} + \frac{1}{2} [\frac{π}{2} + 0]$

$I_{1} = \frac{3 π}{4}$

Let $I_{2} = \int_{0}^{\frac{π}{2}} x^{2} c o s x d x$

$I_{2} = {[x^{2} (s i n x) - \int 2 x \int c o s x d x]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} (s i n x) - 2 \int x s i n x]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} s i n x - 2 (x (- c o s x) + \int c o s x)]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} s i n x - 2 (- x c o s x + s i n x)]}_{0}^{\frac{π}{2}}$

$I_{2} = (\frac{π^{2}}{4} - 2)$

Put l₁ and l₂ in (1)

$\frac{π^{2}}{4} - 2 + \frac{3 π}{4}$

$\frac{π^{2}}{4} + \frac{3 π}{4} - 2$

$\frac{π}{4} (π + 3) - 2$

α = 3

If $| \vec{a} | = 1$ , $| \vec{b} | = 4$ $\vec{a} \cdot \vec{b} = 2$ and $\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$ Then the angle between $\vec{b}$ and $\vec{c}$ is

alok kumar singh

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)

$\vec{c} \cdot \vec{c} = 4 {| \vec{a} \times \vec{b} |}^{2} + 9 {| \vec{b} |}^{2}$

${| \vec{c} |}^{2} = 4 [{| \vec{a} |}^{2} \cdot {| \vec{b} |}^{2} - {(\vec{a} \cdot \vec{b})}^{2}] + 9 {| \vec{b} |}^{2}$

${| \vec{c} |}^{2} = 4 [(1) {(4)}^{2} - (4)] + 9 (16)$

${| \vec{c} |}^{2} = 4 [12] + 144$

${| \vec{c} |}^{2} = 48 + 144$

${| \vec{c} |}^{2} = 192$

$c o s θ = \frac{\vec{b} \cdot \vec{c}}{| \vec{b} | | \vec{c} |}$

$c o s θ = \frac{- 48}{\sqrt[]{192} \cdot 4}$

$c o s θ = \frac{- 48}{8 \sqrt[]{3} \cdot 4}$

$c o s θ = \frac{- 3}{2 \sqrt[]{3}}$

$c o s θ = \frac{- \sqrt[]{3}}{2}$ $θ = c o s^{- 1} (\frac{- \sqrt[]{3}}{2})$

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

alok kumar singh

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

2a + 5b + g – 15 = 0

Also, P lie on line

a + 1 = 2λ

b – 2 = 5λ

g – 4 = λ

2 (2λ – 1) + 5 (5λ + 2) + λ + 4 – 15 = 0

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

$λ = \frac{1}{10}$

a + b + g = (2λ – 1) + (5λ + 2) + (λ + 4)

$= 8 λ + 5 = \frac{8}{10} + 5 = 5.8$

If (α, β, γ) is mirror image of the point (2, 3, 4) with respect to the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ Then 2α + 3β + 4γ is

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Take $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

$\vec{A B}$ = (α − 2) $\hat{i}$ + (β − 3) $\hat{j}$ + (γ − 4) $\hat{k}$

Now,

(α − 2) ⋅ 2 + (β − 3) ⋅3 + (γ − 4) ⋅ 4 = 0

2α − 4 + 3β − 9 + 4γ −16 = 0

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Let $λ$ be an integer. IF the shortest distance between the lines $x - λ = 2 y - 1 = - 2 z$ and $x = y + 2 λ = z - λ i s \frac{\sqrt[]{7}}{2 \sqrt[]{2}}$ , then the value of $| λ | i s . . . . . . . . . . . . . . . . . . .$

Vishal Baghel

$L_{1} = \frac{x - λ}{1} = \frac{y - \frac{1}{2}}{\frac{1}{2}} = \frac{z}{- \frac{1}{2}}$

$∴ S D = \frac{| - 2 λ + 3 (2 λ + \frac{1}{2}) + λ |}{\sqrt[]{14}} = \frac{| 5 λ + \frac{3}{2} |}{\sqrt[]{14}}$

$5 λ + \frac{3}{2} = - \frac{7}{2} \Rightarrow 5 λ = - 5 \Rightarrow λ = - 1$

$∴ | λ | = 1$

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24. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$

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24. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3i^+5j^−6k^

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24. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$