Maths

Contributor-Level 10

Total number of possible relation = $2^{n^{2}} = 2^{4} = 16$ $^{^{}}^{}$

Favourable relations = $?, {(x, x)}, {(y, y)}$

${(x, x), (y, y)}$

${(x, x), (y, y), (x, y), (y, x)}$

Probability = $\frac{5}{16}$

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2 months ago

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x – 3y + 5z = 8. If the mirror image of the point $(2, - \frac{1}{2}, 2)$ in the rotated plane in B(a, b, c), then:

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Contributor-Level 9

Consider the equation of plane,

$P : (2 x + 3 y + z + 20) + λ (x - 3 y + 5 z - 8) = 0$

$?$ Plane P is perpendicular to 2x + 3y + z + 20 = 0

So, $4 + 2 λ + 9 - 9 λ + 1 + 5 λ = 0$

0 $λ = 7$

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$(2, - \frac{1}{2}, 2)$

In plane P is (a, b, c) then

$\frac{a - 2}{1} = \frac{b + \frac{1}{2}}{- 2} = \frac{c - 2}{4}$

and $(\frac{a + 2}{2}) - 2 (\frac{b - \frac{1}{2}}{2}) + 4 (\frac{c + 2}{2}) = 4$

clearly

$a = \frac{4}{3}, b = \frac{5}{6} a n d c = - \frac{2}{3}$

So, a : b : c = 8 : 5 : -4

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2 months ago

Maths Class 11th

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2 a n d l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2} a n d l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4} i s :$

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Maths Ncert Solutions Maths class 11th

Contributor-Level 9

$? l_{1} a n d l_{2}$ are perpendicular, so

$3 * 1 + (- 2) (\frac{α}{2}) + 0 * 2 = 0$

⇒a = 3

Now angle between $l_{2} a n d l_{3}$ ,

$c o s θ = \frac{1 (- 3) + \frac{α}{2} (- 2) + 2 (4)}{\sqrt[]{1 + \frac{α^{2}}{4} + 4} . \sqrt[]{9 + 4 + 16}}$

$\Rightarrow c o s θ = \frac{\frac{2}{29}}{2} \Rightarrow θ = c o s^{- 1} (\frac{4}{29}) = s e c^{- 1} (\frac{29}{4})$

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2 months ago

Let the normal at the point P on the parabola y² = 6x pass through the point (5, -8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is:

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Contributor-Level 9

Let P (at², 2 at) where

$a = \frac{3}{2}$

T : yt = x + at² so point Q is

$(- a, a t - \frac{a}{t})$

N : y = -tx + 2at + at³ passes through (5, -8)

$- 8 = - 5 t + 3 t + \frac{3}{2} t^{3}$

$3 t^{3} - 4 t + 16 = 0$

⇒ t = -2

So ordinate of point Q is $- \frac{9}{4}$

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2 months ago

Maths Vector Algebra

Let A, B, C be three points whose position vectors respectively are

$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$

$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in R$

$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$

If a is the smallest positive integer for which $\vec{a}, \overset{\leftarrow}{b}, \vec{c}$ are noncollinear, then the length of the median, in $Δ A B C,$ through A is:

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Contributor-Level 10

Mid point of BC is $\frac{1}{2} (5 \hat{i} + (α - 2) \hat{j} + 9 \hat{k})$ $\frac{}{}$

$\bar{A B} = \hat{i} + (α - 4) \hat{j} + \hat{k}$

$\bar{A C} = \hat{i} + (- 2 - α) \hat{j} + \hat{k}$

For = 1, $\bar{A B}$ and $\bar{A C}$ will be collinear. So for non collinearity

= 2

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2 months ago

Maths Class 11th

Let C be a circle passing through the points A(2, -1) and B(3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle ${(x - 5)}^{5} + {(y - 1)}^{2} = \frac{13}{2}$ , then r² is equal to:

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Contributor-Level 9

Equation of perpendicular bisector of AB is

$y - \frac{3}{2} = - \frac{1}{5} (x - \frac{5}{2}) \Rightarrow x + 5 y = 10$

Solving it with equation of given circle

${(x - 5)}^{2} + {(\frac{10 - x}{5} - 1)}^{2} = \frac{13}{2}$

$\Rightarrow x - 5 = \pm \frac{5}{2} \Rightarrow x = \frac{5}{2} o r \frac{15}{2}$

But $x \neq \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$(\frac{15}{2}, \frac{1}{2})$

Now,

$r^{2} = {(\frac{15}{2} - 2)}^{2} + {(\frac{1}{2} + 1)}^{2} = \frac{65}{2}$

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2 months ago

Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y +2z = 16. Let T be a plane passing through the point Q and contains the line $\vec{r} = - \hat{k} + λ (\hat{i} + \hat{j} + 2 \hat{k}), λ \in R .$ Then, which of the following points lies on T?

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Contributor-Level 10

$\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 1}{2} = \frac{- 2 (1 + 4 + 2 - 16)}{1 + 2^{2} + 2^{2}}$

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$| \begin{matrix} x & y & z + 1 \\ 3 & 6 & 6 \\ 1 & 1 & 2 \end{matrix} | = 0$ a

=> 2x – z = 1

option (B) satisfies.

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2 months ago

The distance of the origin from the centroid of the triangle whose two sides have the equations x – 2y + 1 = 0 and 2x – y – 1 = 0 and whose orthocenter is $(\frac{7}{3}, \frac{7}{3}) i s :$

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Contributor-Level 10

Let AB $\equiv x - 2 y + 1 = 0$

AC $\equiv x - 2 y + 1 = 0$

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through

orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

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2 months ago

Let $\frac{x - 2}{3} = \frac{y + 1}{- 2} = \frac{z + 3}{- 1}$ lie on the plane px – qy + z = 5, for some $p, q \in R .$ The shortest distance of the plane form the origin is:

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Contributor-Level 10

Line $⊥$ to the normal

=>3p + 2q – 1 = 0

$(2, - 1, - 3)$ lies in the plane 2p + q = 8

From here p = 15, q = -22

Equation of plane 15x – 22y + z – 5 = 0

Distance from origin = $| \frac{5}{\sqrt[]{1 5^{2} + {(- 22)}^{2} + 1^{2}}} | = \sqrt[]{\frac{5}{142}}$

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New answer posted

2 months ago

Maths Class 11th

Let a triangle ABC be inscribed in the circle $x^{2} - \sqrt[]{2} (x + y) + y^{2} = 0$ such that $∠ B A C = \frac{π}{2} .$ If the length of side AB is $\sqrt[]{2},$ then the area of the $Δ A B C$ is equal to:

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