Ncert Solutions Maths class 12th

Let be the plane passing through the point (1, 2, 3) and the line of intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + 4 \hat{k}) = 16 a n d \vec{r} \cdot (- \hat{i} + \hat{j} + \hat{k}) = 6 .$

Then which of the following points does NOT lie on P?

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Equation of required plane

$(x + y + 4 z - 16) + λ (- x + y + z - 6) = 0$ it passes (1, 2, 3)

$\Rightarrow - 1 + λ (- 2) = 0$

$\Rightarrow λ = - \frac{1}{2}$

$∴$ Equation of plane

$(1 - λ) x + (1 + λ) y + (4 + λ) z - 16 - 6 λ = 0$

$\frac{3}{2} x + \frac{1}{2} y + \frac{7}{2} z - 13 = 0$

$\Rightarrow 3 x + y + 7 z = 26$

(4, 2, 2) not satisfying the plane

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New answer posted

2 months ago

Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $(\vec{a} * \vec{b}) . (\vec{b} * \vec{c}) + (\vec{b} * \vec{c}) . (\vec{c} * \vec{a}) + (\vec{c} * \vec{a}) . (\vec{a} * \vec{b})$ = 168, then $| \vec{a} | + | \vec{b} | + | \vec{c} |$ is equal to

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Given : $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = c o s θ (s a y)$

$| \vec{a} | | \vec{b} | | \vec{c} | = 14$

$(\vec{a} * \vec{b}) \cdot (\vec{b} * \vec{c})$

$= \vec{a} \cdot [(\vec{b} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{c}]$

$= (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) - {| \vec{b} |}^{2} \vec{a} \cdot \vec{c}$

$= | \vec{a} | {| \vec{b} |}^{2} | \vec{c} | (c o s^{2} θ - c o s θ) = 14 | \vec{b} | (c o s^{2} θ - c o s θ)$

Similarly, $(\vec{b} * \vec{c}) \cdot (\vec{c} * \vec{a})$

= $| \vec{b} | {| \vec{c} |}^{2} | \vec{a} | (c o s^{2} θ - s i n θ) = 14 | \vec{c} | (c o s 2 θ - s i n θ) & (\vec{c} * \vec{a}) \cdot (\vec{a} * \vec{b})$

$= | \vec{c} | {| \vec{a} |}^{2} | \vec{b} | (c o s^{2} θ - s i n θ) = 14 | \vec{a} | (c o s^{2} θ - s i n θ)$

Given : 14 $(c o s^{2} θ - c o s θ) (| \vec{a} | + | \vec{b} | + | \vec{c} |) = 168$

$| \vec{a} | + | \vec{b} | + | \vec{c} | = \frac{12}{c o s^{2} θ - c o s θ} = \frac{12}{\frac{1}{4} - (- \frac{1}{2})} = \frac{12}{\frac{3}{4}} = 16$

Given : $\vec{a}, \vec{b}, \vec{c}$ are coplanar & pair wise equal angle.

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New answer posted

2 months ago

Let $S = {z = x + i y | z - 1 + i | \geq | z |, | z | < 2, | z + i | = | z - 1 |} .$ Then the set of all values of x, for which $w = 2 x + i y \in S$ for some $y \in R,$ is

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$| z - 1 + i | \geq | z |$

$| z + i | = | z - 1 |$

W = (2x, y) = (α, y)

Let S represent the line segment AB

For 'B'

x² + y² = 4

x = y

x² = 2

$x = \pm \sqrt[]{2}$

$B (- \sqrt[]{2}, \sqrt[]{2})$

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

W (2x, y) lies on AB

$- \sqrt[]{2} < 2 x \leq \frac{1}{2}$

$(| z | < 2)$

$- \frac{1}{\sqrt[]{2}} < x \leq \frac{1}{4}$

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2 months ago

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The balls so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is

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$\begin{matrix} 3 R & 2 R & 4 B \\ 5 B & 3 W & 2 W \end{matrix}$

I II

Let

E₁ : a red ball is transferred from I to II

E₂ : a black is transferred from I to II

E₃ :a white transferred from I to II

E : a black ball is drawn from 2nd bag after a ball from I to II was transferred.

$P (\frac{E_{1}}{E}) = \frac{P (E_{1} \cap E)}{P (E)}$

$P (E) = P (E_{1} \cap E) + P (E_{2} \cap E) + P (E_{3} \cap E)$

= $P (E_{1}) \cdot P (\frac{E}{E_{1}}) + . . . . + . . . . .$

= $\frac{3}{10} \cdot \frac{5}{10} + \frac{4}{10} \cdot \frac{6}{10} + \frac{3}{10} \cdot \frac{5}{10} = \frac{54}{100}$

$P (E_{1} / E) = \frac{15 / 100}{54 / 100} = \frac{5}{18}$

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New answer posted

2 months ago

For I(x) = $\int \frac{s e c^{2} x - 2022}{s i n^{2022} x} d x, i f I (\frac{π}{4}) = 2^{1011}, t h e n$

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I(x) = $\int \frac{s e c^{2} x - 2022}{s i n^{2022} x} d x$

$= \int s i n^{- 2022} x \cdot s e c^{2} x d x - \int 2022 s i n^{- 2022} x d x$

$= s i n^{- 2022} x \cdot t a n x - \int (- 2022) s i n^{- 2023} x \cdot c o s x \cdot t a n x d x - \int 2022 s i n^{- 2022} x d x$

$= t a n x \cdot s i n^{- 2022} x + 2022 \int s i n^{- 2022} - 2022 \int s i n^{- 2022} x d x$

I(x) = $t a n x \cdot s i n^{- 2022} x + c$

Given, $I (\frac{π}{4}) = 2^{1011}$

$2^{1011} = \frac{1}{{(\frac{1}{\sqrt[]{2}})}^{2022}} + c \Rightarrow c = 0$

$I (x) = \frac{t a n x}{s i n^{2022} x}, I (\frac{π}{3}) = \frac{\sqrt[]{3}}{(\frac{\sqrt[]{3}}{2})} = \frac{2^{2022}}{{(\sqrt[]{3})}^{2021}}$

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2 months ago

Let the function $f (x) = {\frac{l o g_{e} (1 + 5 x) - l o g_{e} (1 + α x)}{\begin{matrix} x & 10 \end{matrix}} \begin{matrix} ; i f x \pm 0 & ; i f x = 0 \end{matrix} b e c o n t i n u o u s a t x = 0 .$ Then α is equal to

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New answer posted

2 months ago

Let $l_{1}$ be the line in xy-plane with x and y intercepts $\frac{1}{8} a n d \frac{1}{4 \sqrt[]{2}}$ respectively, and $l_{2}$ be the line in zx-plane with x and z intercepts $- \frac{1}{8} a n d - \frac{1}{6 \sqrt[]{3}}$ respectively. If d is the shortest distance between the line $l_{1} a n d l_{2},$ then d^-2 is equal to……………

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$l_{1} :$

$\vec{r} = \frac{1}{8} \hat{i} + λ (- \frac{1}{8}, \frac{1}{4 \sqrt[]{2}}, 0)$

$\vec{r} = - \frac{1}{8} \hat{i} + μ (\frac{1}{8}, 0, - \frac{- 1}{6 \sqrt[]{3}})$

$\vec{n} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{- 1}{8} & \frac{1}{4 \sqrt[]{2}} & 0 \\ \frac{1}{8} & 0 & \frac{- 1}{6 \sqrt[]{3}} \end{matrix} |$

$= (- \frac{1}{24 \sqrt[]{6}}, \frac{- 1}{48 \sqrt[]{3}}, - \frac{1}{32 \sqrt[]{2}})$

$d = p r o j e c t i o n o f A \vec{C} o n \vec{n} = (- \frac{2}{8}, 0, 0) . \frac{(4, 2 \sqrt[]{2}, 3 \sqrt[]{3})}{\sqrt[]{16} + 8 + 27} = \frac{- 1}{\sqrt[]{16 + 8 + 27}}$

d²⁼ $\frac{1}{51}$

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New answer posted

2 months ago

Let $f (x) = | (x - 1) (x^{2} - 2 x - 3) | + x - 3, x \in R .$ If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to………….

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Contributor-Level 10

$f (x) = | (x - 1) (x + 1) (x - 3) | + x - 3$

For $x \in [- 1, 1] \cup [3, \infty)$

y = x² (x – 3)

$\frac{d y}{d x} = 2 x (x - 3) + x^{2} = 3 x^{2} - 6 x = 3 x (x - 2)$

Number of max = 2

Number of min = 1

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New answer posted

2 months ago

Let $\vec{b} = \hat{i} + \hat{j} + λ \hat{k}, λ \in R . I f \vec{a}$ is a vector such that $\vec{a} * \vec{b} = 13 \hat{i} - \hat{j} - 4 \hat{k} a n d \vec{a} . \vec{b} + 21 = 0,$ then $(\vec{b} - \vec{a}) . (\hat{k} - \hat{j}) + (\vec{b} + \vec{a}) (\hat{i} - \hat{k})$ is equal to

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Contributor-Level 10

$\vec{a} * \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 1 & 1 & λ \end{matrix} |$

$= (λ a_{2} - a_{3}, a_{3} - a_{1} λ, a_{1} - a_{2})$

$= (13, - 1, - 4)$

$\vec{a} . \vec{b} = - 21$

$\Rightarrow a_{1} + a_{2} + a_{3} λ = - 21$

$λ a_{2} - a_{3} = 13 \Rightarrow λ = \frac{13 + a_{3}}{a_{2}} = \frac{a_{3} + 1}{a_{1}}$

$\vec{b} - \vec{a} = (3, - 1, 10)$

$\vec{b} + \vec{a} = (- 1, 3, - 4)$

= 11 + 3 + 14

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New answer posted

2 months ago

The value of b > 3 for which $12 \int_{3}^{b} \frac{1}{(x^{2} - 1) (x^{2} - 4)} d x = l o g_{e} (\frac{49}{40}),$ is equal to………

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