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Class 11th Maths

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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Raj Pandey

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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Class 11th Ncert Solutions Maths class 11th

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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Raj Pandey

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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Class 11th 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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Vishal Baghel

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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Class 11th Maths

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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Raj Pandey

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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Class 11th Chemistry Organic Chemistry

In $\overset{1}{C} H_{2} = \overset{2}{C} = \overset{3}{C} H - \overset{4}{C} H_{3}$ molecule, the hybridization of carbon 1,2,3 and 4 respectively are:

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alok kumar singh

Contributor-Level 10

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Class 11th Maths

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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Vishal Baghel

Contributor-Level 10

${(x - \frac{1}{\sqrt[]{2}})}^{2} + {(y - \frac{1}{\sqrt[]{2}})}^{2} = 1$

here AB = $\sqrt[]{2}$ , BC = 2, AC = $\sqrt[]{2}$

area = $\frac{1}{2} * \sqrt[]{2} * \sqrt[]{2} = 1$

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Class 11th Chemistry

Calgon is used for water treatment. Which of the following statement is NOT true about Calgon?

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alok kumar singh

Contributor-Level 10

It does not contains 2^nd most abundant element by weight in earth crust because that is Si Calgon -> Na₂ [Na₄ (PO₃)₆]

$\to_{}^{W a t e r s o l u b l e} 2 N a^{+} {[N a_{4} {(P O_{3})}_{6}]}^{2 -}$

$\to_{}^{C a^{2 +}} 2 N a^{+} {[N a_{2} C a {(P O_{3})}_{6}]}^{2 -}$

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Class 11th Maths

Let P : y² = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of $\frac{π}{4}$ with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is

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Vishal Baghel

Contributor-Level 10

Tangents making angle $\frac{π}{4}$ $\frac{}{}$ with y = 3x + 5.

$t a n \frac{π}{4} = | \frac{m - 3}{1 + 3 m} | \Rightarrow m = - 2, \frac{1}{2}$

So, these tangents are $⊥$ . So ASB is a focal chord.

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Class 11th Maths

Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $Δ P Q R$ is:

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