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Three Dimensional Geometry

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5 months ago

Three Dimensional Geometry Class 12th

Let $S = {(x, y) \in N * N : 9 {(x - 3)}^{2} + 16 {(y - 4)}^{2} \leq 144}$ and $T = {(x, y) \in R * R : {(x - 7)}^{2} + {(y - 4)}^{2} \leq 36} .$ Then $n (S \cap T)$ is equal to

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Payal Gupta

Contributor-Level 10

$\frac{{(x - 3)}^{2}}{16} + \frac{{(y - 4)}^{2}}{9} \leq 1, x, y \in N, {(x - 7)}^{2} + {(y - 4)}^{2} \geq 36, x, y \in R$

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

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5 months ago

Three Dimensional Geometry Class 12th

Let $P (- 2, - 1, 1)$ and $Q (\frac{56}{17}, \frac{43}{17}, \frac{111}{17})$ be the vertices of the rhombus PQRS. If the direction ratios of the diagonal RS are a, -1, b where both a and b are integers of minimum absolute values, then α² + β² is equal to……….

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Vishal Baghel

Contributor-Level 10

$R S \equiv (α, -, 1 β)$

$D R o f P Q \equiv (\frac{56}{17} + 2, \frac{43}{17} + 1, \frac{111}{17} - 1)$

$\equiv (\frac{90}{17}, \frac{60}{17}, \frac{94}{17})$

$90 α + 94 β = 60$

$\begin{array}{l} β = - 15, α = - 15 \\ α^{2} + β^{2} = 450 \end{array}$

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5 months ago

Three Dimensional Geometry Class 12th

Let the lines

L₁: $\vec{r} = λ (\vec{i} + 2 \hat{j} + 3 \hat{k}), λ \in R$

L₂ : $\vec{r} = (\hat{i} + 3 \hat{j} + \hat{k}) + μ (\hat{i} + \hat{j} + \hat{5}); μ \in R,$

intersects at the point. S. If a plane ax + by – z + d = 0 passes through S and is parallel to both lines L₁ and L₂, then the value of a + b + d is equal to…………

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Vishal Baghel

Contributor-Level 10

$λ = 1 + μ$

$\begin{array}{l} \underline{2 λ = 3 λ μ} \\ λ = 2 \to S (2, 4, 6) \end{array}$

$μ = 1$

$\vec{n} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & 1 & 5 \end{matrix} | = (7, - 2, - 1)$

7x – 2y – z + d = 0

(2, 4, 6) Þ d = -14 + 8 + 6

d = 0

7x – 2y – z = 0

7 – 2 = 5

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5 months ago

Three Dimensional Geometry Class 12th

Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, -1, -1) parallel to the line PQ meets the plane S at R, then QR² is equal to:

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Vishal Baghel

Contributor-Level 10

$l i m P Q : \frac{x - 1}{1} = \frac{y - 0}{1} = \frac{3 - 1}{1} = λ$

$M (1 + λ, λ, 1 + λ)$

x + y + z = 5

$\Rightarrow λ = 1$

M (2, 1, 2)

Q (3, 2, 3)

$L : \frac{x - 1}{1} = \frac{y + 1}{1} = \frac{z + 1}{1} = t$

R (3, 1, 1), QR² = 1 + 4 = 5

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5 months ago

Three Dimensional Geometry Class 12th

$\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} a_{i} > 0, i = 1, 2, 3$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $\vec{a}$ on the vector $3 \hat{i} + 4 \hat{j} b e 7 .$ Let $\vec{b}$ be a vector obtained by rotating with and x-axis are coplanar, then projection of a vector $\vec{b}$ on $3 \hat{i} + 4 \hat{j}$ is equal to:

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Vishal Baghel

Contributor-Level 10

$\vec{a} = (t, t, t)$

$\frac{3 t + 4 t}{5} = 7 \Rightarrow t = 5$

$\vec{a} = (5, 5, 5)$

$\vec{b} = l \vec{a} + m \hat{i}$

$= (5 l + m, 5 l, 5 l)$

$\frac{\vec{b} . (3, 4, 0)}{5} = \frac{\pm \frac{5}{\sqrt[]{2}} (- 6 + 4 + 0)}{5} = \pm (\sqrt[]{2})$

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5 months ago

Three Dimensional Geometry Class 12th

If the shortest distance between the lines $\vec{r} = (- \hat{i} + 3 \hat{k}) + λ (\hat{i} - a \hat{j}) a n d \vec{r} = (- \hat{j} + 2 \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})$ is $\sqrt[]{\frac{2}{3}}$ , then the integral value of 'a' is equal to __________.

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alok kumar singh

Contributor-Level 10

${\vec{b}}_{1} * {\vec{b}}_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - a & 0 \\ 1 & - 1 & 1 \end{matrix} |$ = $- a \hat{i} - \hat{j} + (a - 1) \hat{k}$

${\vec{a}}_{1} - {\vec{a}}_{2} = - \hat{i} + \hat{j} + \hat{k}$

Shortest distance = $| \frac{({\vec{a}}_{1} - {\vec{a}}_{2}) . ({\vec{b}}_{1} * {\vec{b}}_{2})}{| {\vec{b}}_{1} * {\vec{b}}_{2} |} |$

$= \frac{2 (a - 1)}{\sqrt[]{a^{2} + 1 + {(a + 1)}^{2}}} = \sqrt[]{\frac{2}{3}} = \frac{4 {(a - 1)}^{2}}{a^{2} + 1 + {(a - 1)}^{2}} = \frac{2}{3}$

$12 {(a - 1)}^{2} = 2 (a^{2} + 1) + 2 {(a - 1)}^{2}$

$10 {(a - 1)}^{2} = 2 (a^{2} + 1)$

$5 a^{2} - 10 a + 5 = a^{2} + 1 \Rightarrow 4 a^{2} - 10 a + 4 = 0$

$2 a^{2} - 5 a + 2 = 0$

$(2 a - 1) (a - 2) = 0$

$∴ a = \frac{1}{2}, 2$

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5 months ago

Three Dimensional Geometry Class 12th

Let a line having direction ratios 1, -4, 2 intersect the lines $\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1}$ and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1}$ at the points A and B. Then (AB)² is equal to _________.

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alok kumar singh

Contributor-Level 10

$\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1} = r_{1}$

$A (3 r_{1} + 7, 1 - r_{1}, - 2 + r_{1})$

and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1} = r_{2}$

$B (2 r_{2}, 7 + 3 r_{2}, r_{2})$

$A / q, \frac{3 r_{1} - 2 r_{2} + 7}{1} = \frac{3 r_{2} + r_{1} + 6}{4} = \frac{r_{1} - r_{2} - 2}{2}$

$\Rightarrow r_{1} = - 5, r_{2} = - 3$

$∴ A (- 8, 6, - 7) a n d B (- 6, - 2, - 3)$

AB² = 84

$f (x) = {\begin{cases} | 2 x^{2} - 3 x - 7 |, x \leq - 1 \\ [4 x^{2} - 1], - 1 < x < 1 \\ | x + 1 | + | x - 2 |, x \geq 1 \end{cases}$

f(-1) = 1

f(1) = 3

Hence f(x) will be discontinuous at x = 1 and also 4x² – 1 = 0 , 1 , 2

$\Rightarrow x = \pm \frac{1}{2}, \pm \frac{1}{\sqrt[]{2}}, \pm \frac{\sqrt[]{3}}{2}$

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New answer posted

5 months ago

Three Dimensional Geometry Class 12th

Let the points on the plane P be equidistance from the points (–4, 2, 1) and (2, –2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is

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Payal Gupta

Contributor-Level 10

Let P (x, y, z) be any point on plane P₁ then

${(x + 4)}^{2} + {(y - 2)}^{2} + {(z - 1)}^{2} = {(x - 2)}^{2} + {(y + 2)}^{2} + {(z - 3)}^{2}$

$∴ c o s θ = \frac{| 6 - 2 + 3 |}{14} \Rightarrow θ = \frac{π}{3}$

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New answer posted

5 months ago

Three Dimensional Geometry Class 12th

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5} i s \frac{1}{\sqrt[]{3}},$ then the sum of all possible values of $λ$ is

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Payal Gupta

Contributor-Level 10

S. D. = $\frac{1}{\sqrt[]{3}}$

$b_{1} * b_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & λ \\ 1 & 4 & 5 \end{matrix} | = \hat{i} (15 - 4 λ) + \hat{j} (λ - 10) + \hat{k} (5)$

$\frac{| (- \hat{i} - 2 \hat{j} - 2 \hat{k}) . {(15 - 4 λ) \hat{i} + (λ - 10) \hat{j} + 5 \hat{k}} |}{\sqrt[]{{(15 - 4 λ)}^{2} + {(λ - 10)}^{2} + 25}} = \frac{1}{\sqrt[]{3}}$

$\Rightarrow λ = 16$

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New answer posted

5 months ago

Three Dimensional Geometry Class 12th

Let a triangle be bounded by the lines L₁ : 2x + 5y = 10; L₂ : -4x + 3y = 12 and the line L₃, which passes through the point P(2, 3), intersects L₂ at A and L₁ at B. If the point P divides the line-segment AB, internally in the nation 1 :3, then the area of the triangle is equal to:

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alok kumar singh

Contributor-Level 10

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