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Vector Algebra

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2 months ago

Vector Algebra Class 12th

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\vec{a} * \vec{b} = 4 \vec{c}, \vec{b} * \vec{c} = 9 \vec{a}$ and $\vec{c} * \vec{a} = α \vec{b}, α > 0 .$ If $| \vec{a} | + | \vec{b} | + | \vec{c} | = \frac{1}{36},$ then a is equal to…….

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Vishal Baghel

Contributor-Level 10

Data contradiction.

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2 months ago

Vector Algebra Class 12th

Let $\vec{a} a n d \vec{b}$ be two vectors such that $| \vec{a} + \vec{b} | = {| \vec{a} |}^{2} + 2 {| \vec{b} |}^{2}, \vec{a} . \vec{b} = 3 a n d {| \vec{a} * \vec{b} |}^{2} = 75 .$ Then ${| \overset{\leftrightarrow}{a} |}^{2}$ is equal to

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Payal Gupta

Contributor-Level 10

Given : ${| \vec{a} + \vec{b} |}^{2} = {| \vec{a} |}^{2} + 2 {| \vec{b} |}^{2} & \vec{a} \cdot \vec{b} = 3$

$| \vec{a} | | \vec{b} | c o s θ = 3$

$| \vec{a} | c o s θ = \frac{3}{\sqrt[]{6}} = \sqrt[]{\frac{9}{6}} = \sqrt[]{\frac{3}{2}}$

${| \vec{a} |}^{2} {| \vec{b} |}^{2} s i n^{2} θ = 75$

$| \vec{a} | s i n θ = \sqrt[]{\frac{75}{6}} = \frac{5}{\sqrt[]{2}}$

$| \vec{a} | c o s θ = \sqrt[]{\frac{3}{2}}$

${| \vec{a} |}^{2} = \frac{25}{2} + \frac{3}{2} = \frac{28}{2} = 14$

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New answer posted

2 months ago

Vector Algebra Class 12th

Let θ be the angle between the vectors $\vec{a} a n d \vec{b},$ where $| \vec{a} | = 4, | \vec{b} | = 3$ and $θ \in (\frac{π}{4}, \frac{π}{3})$ . Then ${| (\vec{a} - \vec{b}) * (\vec{a} + \vec{b}) |}^{2} + 4 {(\vec{a} . \vec{b})}^{2}$ is equal to…………

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Vishal Baghel

Contributor-Level 10

${| 2 (\vec{a} * \vec{b}) |}^{2} + 4 {(\vec{a} . \vec{b})}^{2}$

$4 {| \vec{a} |}^{2} {| \vec{b} |}^{2}$

= 4 * 16 * 9 = 576

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New answer posted

2 months ago

Vector Algebra Class 12th

Let $\hat{a}, \hat{b}$ and be unit vectors. If $\hat{c}$ be vector such that the angle between $\hat{a} a n d \vec{c} i s \frac{π}{12},$ and $\hat{b} = \vec{c} + 2 (\vec{c} * \hat{a}),$ then ${| 6 \vec{c} |}^{2}$ is equal to :

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alok kumar singh

Contributor-Level 10

$| \hat{a} | = 1, | \hat{b} | = 1$

${| \hat{b} |}^{2} = {| \vec{c} + 2 (\vec{c} * \hat{a}) |}^{2}$

$\Rightarrow 1 = {| \vec{c} |}^{2} + 4 {| \vec{c} |}^{2} {(\frac{\sqrt[]{3} - 1}{2 \sqrt[]{2}})}^{2}$

${| \vec{c} |}^{2} [1 + 4 * \frac{{(\sqrt[]{3} - 1)}^{2}}{8}] = {| \vec{c} |}^{2} (3 - \sqrt[]{3})$

$∴ {| \vec{c} |}^{2} = \frac{1}{3 - \sqrt[]{3}} = \frac{3 + \sqrt[]{3}}{6}$

$\Rightarrow {| 6 \vec{c} |}^{2} = 6 (3 + \sqrt[]{3})$

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New answer posted

2 months ago

Vector Algebra Class 12th

Let S = ${z \in C : | z - 3 | \leq 1 a n d z (4 + 3 i) \leq 24} .$ If $α + i β$ is the point in S which in closest to 4i, then 25 $(α + β)$ is equal to

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Payal Gupta

Contributor-Level 10

Circle $| z - 3 | \leq 1 \Rightarrow {(x - 3)}^{2} + y^{2} \leq 1$

and line $z (4 + 3 i) + \bar{z} (4 - 3 i) \leq 24$

$\Rightarrow 4 x - 3 y \leq 12 ∴ s l o p e = t a n θ = \frac{4}{3}$

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New answer posted

2 months ago

Vector Algebra Class 12th

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that $| (\hat{a} + \hat{b}) + 2 (\hat{a} * \hat{b}) |$ = 2. If $θ \in (0, π)$ is the angle between $\hat{a} a n d \hat{b}$ , then among the statements:

$(S 1) : 2 | \hat{a} * \hat{b} | = | \hat{a} - \hat{b} |$

(S1) : The projection of $\hat{a} o n (\hat{a} + \hat{b}) i s \frac{1}{2}$

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Payal Gupta

Contributor-Level 10

$| \vec{a} + \vec{b} + 2 (\vec{a} * \vec{b}) | = 2, θ \in (0, π)$

squaring on both sides, we get = $\frac{2 π}{3}$

where θ is angle between $\hat{a} a n d \hat{b} .$

$2 | \hat{a} * \hat{b} | = \sqrt[]{3} = | \hat{a} - \hat{b} |, S_{1}$ is correct.

and projection of $\hat{a} o n \hat{a} + \hat{b} = | \frac{\hat{a} . (\hat{a} + \hat{b})}{| \hat{a} + \hat{b} |} | = \frac{1}{2}$

So (S₂) is correct.

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New answer posted

2 months ago

Vector Algebra Class 12th

$\underset{x \to \infty}{l i m} (\frac{n^{2}}{(n^{2} + 1) (n + 1)} + \frac{n^{2}}{(n^{2} + 4) (n + 2)} + \frac{n^{2}}{(n^{2} + 9) (n + 3)} + . . . + \frac{n^{2}}{(n^{2} + n^{2}) (n + n)}) i s e q u a l t o$

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Payal Gupta

Contributor-Level 10

S = $\underset{n \to \infty}{L t} \sum_{r = 1}^{n} \frac{n^{2}}{(n^{2} + r^{2}) (n + r)}$

$\frac{π}{8} + \frac{1}{4} l n 2$

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New answer posted

2 months ago

Vector Algebra Class 12th

Let the lope of the tangent to a curve y = f(x) at (x,y) be given by 2 tan x (cos x – y). If the curve passes through the point $(\frac{π}{4}, 0),$ then the value of $\int_{0}^{π / 2} y d x$ is equal to:

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alok kumar singh

Contributor-Level 10

$\frac{d y}{d x} + 2 y t a n x = 2 s i n x$

$I . F . = e^{2 \int t a n x d x} = s e c^{2} x$

$∴$ Solution y sec²x = $2 \int s i n x s e c^{2} x d x = 2 \int s e c x t a n x d x$

y sec² x = 2sec x + c

y = 2 cos x + c cos²x passes $(\frac{π}{4}, 0) = B$ C = $- 2 \sqrt[]{2}$

y = 2 cos x $2 \sqrt[]{2} c o s^{2} x$ $∴ \int_{0}^{π / 2} y d x = 2 - \frac{π}{\sqrt[]{2}}$

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New answer posted

2 months ago

Vector Algebra Class 12th

Let $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k} a n d \vec{c}$ be a vector such that $\vec{a} + (\vec{b} * \vec{c}) = \vec{0} a n d \vec{b} . \vec{c} = 5 .$ Then, the value of $3 (\vec{c} . \vec{a})$ is equal to

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Payal Gupta

Contributor-Level 10

Data contradiction.

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

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New answer posted

2 months ago

Vector Algebra Class 12th

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 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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Payal Gupta

Contributor-Level 10

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

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

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

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

= 2

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