Maths

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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alok kumar singh

Contributor-Level 10

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

9 months ago

Let a circle C in complex plane pass through the points z₁ = 3 + 4i, z₂ = 4 + 3i and z₃ = 5i. If $z (\neq z_{1})$ is a point on C such that the line through z and z₁ is perpendicular to the line through z₂ and z₃, then arg(z) is equal to:

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Vishal Baghel

Contributor-Level 10

$\frac{2}{- 4} = \frac{- 1}{2}$

y – 4 = 2 (x – 3)

y = 2x – 2

x² + (2x – 2)² = 25

$5 x^{2} - 8 x - 21 = 0$

$z (\frac{- 7}{5}, \frac{- 24}{5})$

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New answer posted

9 months ago

Maths Differential Equations

Let x = 2t, $y = \frac{t^{2}}{3}$ be a conic. Let S be the focus and B the point on the axis of the conic such that $S A ⊥ B A,$ where A is any point on the conic. If k is the ordinate of the centroid of the $Δ S A B,$ then $\underset{t \to 1}{l i m}$ k is equal to:

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Vishal Baghel

Contributor-Level 10

x = 2t $A (2 t, \frac{t^{2}}{3})$ $\frac{^{}}{}$

$y = \frac{t^{2}}{3}$ S (0, 3)

${\frac{}{}}^{}$ $3 y = {(\frac{x}{2})}^{2}$ $B (0, λ)$

$3 k = \frac{t^{2}}{3} + 3 + λ = \frac{2 t^{2}}{3} + 3 - \frac{12 t^{2}}{9 - t^{2}}$

$\underset{t \to 1}{l i m} 3 k = \frac{2}{3} + 3 - \frac{12}{8} = 3 - \frac{5}{6} = \frac{13}{6}$

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New answer posted

9 months ago

If the solution curve y = y(x) of the differential equation y²dx + (x² – xy + y²)dy = 0, which passes through the point (1, 1) and intersects the line y = $\sqrt[]{3} x$ at the point $(α, \sqrt[]{3} α),$ then value of $l o g_{e} (\sqrt[]{3} α)$ is equal to:

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Vishal Baghel

Contributor-Level 10

$y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$

$\frac{d x}{d y} + \frac{x^{2} - x y + y^{2}}{y^{2}} = 0 \Rightarrow \frac{d x}{d y} + {(\frac{x}{y})}^{2} - (\frac{x}{y}) + 1 = 0$

Put x = vy $\Rightarrow v + y \frac{d v}{d y} + v^{2} - v + 1 = 0$

$\Rightarrow \frac{y d v}{d y} + v^{2} + 1 = 0$

$\Rightarrow \frac{π}{6} + l n | y | = \frac{π}{4}$

$l n | y | = \frac{π}{12}$

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New answer posted

9 months ago

The statement $(p \Rightarrow q) \lor (p \Rightarrow r)$ is NOT equivalent to

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Maths Relations and Functions

Contributor-Level 10

$(A) \sim (p \land (\sim r)) \lor q = (\sim p) \lor (r) \lor q$

(B) $q \lor (- r \lor p)$

(C) $(\sim p) \lor (q \lor r)$

(D) $(\sim p \lor q) \lor r$

Using Venn diagram we get B as the correct option.

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9 months ago

The domain of the function f(x) = $s i n^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7}) i s :$

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Contributor-Level 10

$- 1 \leq \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$

$\Rightarrow 0 \leq \frac{2 x^{2} - x + 9}{x^{2} + 2 x + 7} & \frac{- 5 x - 5}{x^{2} + 2 x + 7} \leq 0$

$x \in R & - 1 \leq x < \infty$

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New answer posted

9 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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Contributor-Level 10

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

9 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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Contributor-Level 10

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

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

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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New answer posted

9 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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Contributor-Level 10

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

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

9 months ago

If (2, 3, 9), (5, 2, 1), $(1, λ, 8)$ and $(λ, 2, 3)$ are coplanar, then the product of all possible values of $λ$ is

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