Vector Algebra

Let three vectors $\vec{a}, \vec{b} a n d \vec{c}$ such that $\vec{a} * \vec{b} = \vec{c}, \vec{b} * \vec{c} = \vec{a} a n d | \vec{a} | = 2 .$ Then which one of the following is not true?

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Contributor-Level 10

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

$\vec{b} * \vec{c} = \vec{a} - (i i) | \vec{a} | = 2$

Taking dot product with $\vec{c} & \vec{a}$ respectively in (i) & (ii) we have

$[\vec{c} \vec{a} \vec{b}] = {| \vec{c} |}^{2}$

Again (i) & (ii) $\vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0$ Þ

Opt 1. Projection of $\vec{a} o n \vec{b} * \vec{c} = \frac{\vec{a} . (\vec{b} * \vec{c})}{| \vec{b} * \vec{c} |} = \frac{4}{2} = 2$

Opt 2. $[\vec{a} \vec{b} \vec{c}] + [\vec{c} \vec{a} \vec{b}] = 2 [\vec{a} \vec{b} \vec{c}] = 8$ (2)

(1) $\vec{b} * (\vec{a} * \vec{b}) = \vec{b} * \vec{c}$ Þ

Opt 3. ${| 3 \vec{a} + \vec{b} - 2 \vec{c} |}^{2} = 9 {| \vec{a} |}^{2} + {| \vec{b} |}^{2} + 4 {| \vec{c} |}^{2} + 2 (3 \vec{a} . \vec{b} - 6 \vec{a} . \vec{c} - 2 \vec{b} . \vec{c})$

Opt 4. $\vec{a} * (\vec{c} * \vec{b} - \vec{b} * \vec{c})$

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New answer posted

a month ago

Let a vector $\vec{a}$ be coplanar with vectors $\vec{b} = 2 \hat{i} + \hat{j} + \hat{k} a n d \vec{c} = \hat{i} - \hat{j} + \hat{k} . I f \vec{a}$ is perpendicular to $\vec{d} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} a n d | \vec{a} | = \sqrt[]{10 .}$ Then a possible value of $[\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{d}] + [\vec{a} \vec{c} \vec{d}]$ is equal to:

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Contributor-Level 10

$\vec{a} = α \vec{b} + β \vec{c} = (2 α + β) i + (α - β) J + (α + β) k$

is perpendicular to $\vec{d} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$

$\Rightarrow 3 (2 α + β) + 2 (α + β) + 6 (α + β) = 0$

->14a + 7b = 0 Þ b = -2a .(i)

$\Rightarrow \vec{a} = 3 α j - α k$

$| \vec{a} | = \sqrt[]{9 α^{2} + α^{2}} = \sqrt[]{10} \Rightarrow | α | = 1 \Rightarrow α \pm 1, f o r α = 1, \vec{a} = 3 \hat{j} - \hat{k}$

for $α = - 1, \vec{a} = - 3 \hat{j} + \hat{k}$

$\vec{d} * \vec{a} = \pm | \begin{matrix} i & j & k \\ 3 & 2 & 6 \\ 0 & 3 & - 1 \end{matrix} |$

$= \pm 42$

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New answer posted

a month ago

For p > 0, a vector ${\vec{v}}_{2} = 2 \hat{i} + (p + 1) \hat{j}$ is obtained by rotating the vector ${\vec{v}}_{1} = \sqrt[]{3} p \hat{i} + \hat{j}$ by an angle q about origin in counter clockwise direction. If tan θ $\frac{(α \sqrt[]{3} - 2)}{(4 \sqrt[]{3} + 3)},$ = then the value of a is equal to……….

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Contributor-Level 10

$? | {\vec{v}}_{1} | = | {\vec{v}}_{2} | \Rightarrow p^{2} - p - 2 = 0 \Rightarrow p - 1, 2$ we take only p = 2 (p > 0)

$∴ c o s θ = \frac{\vec{v_{2}} . \vec{v_{2}}}{| \vec{v_{1}} | | \vec{v_{2}} |} = \frac{4 \sqrt[]{3} + 3}{13} \Rightarrow t a n θ = \frac{6 \sqrt[]{3} - 2}{4 \sqrt[]{3} + 3} = \frac{α \sqrt[]{3} - 2}{4 \sqrt[]{3} + 3} ∴ α = 6$

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New answer posted

2 months ago

Let $\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} a n d \vec{b} = \hat{i} + 2 \hat{j} - \hat{k} .$ Let a vector $\vec{v}$ be in the plane containing $\vec{a} a n d \vec{b} .$ If $\vec{v}$ is perpendicular to the vector $3 \hat{i} + 2 \hat{j} - \hat{k}$ and its projection on $\vec{a}$ is 19 units, then ${| 2 \vec{v} |}^{2}$ is equal to_________.

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Contributor-Level 10

$\vec{v} = l \vec{a} + m \vec{b}$

$= (2 l + m, - l + 2 m, 2 l = m)$

$\vec{v} . (3, 2, - 1) = 0 \Rightarrow l = - 4 m . . . . . . . (i)$

$| \vec{v} . \hat{a} | = 19 \Rightarrow 9 l - 2 m = 57 . . . . . . . . (i i)$

(i) & (ii) Þ l = 6, m = $\frac{- 3}{2}$

$\Rightarrow 2 \vec{v} = (21, - 18, 27)$

${| 2 \vec{v} |}^{2} = 1494$

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New answer posted

2 months ago

Let θ be the acute angle between the tangents to the ellipse $\frac{x}{9} + \frac{y}{1} = 1$ and the circle x² + y² = 3 at their point of intersection in the first quadrant. The tanq is equal to :

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Contributor-Level 10

Point of intersection of

$\frac{x^{2}}{9} + y^{2} = 1 a n d x^{2} + y^{2} = 3 i n t h e 1 s t q u a d r a n t i s (\frac{3}{2}, \frac{\sqrt[]{3}}{2})$

$m_{1} = - \frac{1}{3 \sqrt[]{3}}, m_{2} = - \sqrt[]{3}$

$t a n θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = \frac{2}{\sqrt[]{3}}$

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New answer posted

2 months ago

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $\vec{r}$ satisfies $\vec{a} * {(\vec{r} - \vec{b}) * \vec{a}} + \vec{b} * {(\vec{r} - \vec{b}) * \vec{b}} + \vec{c} * {(\vec{r} - \vec{a}) * \vec{c}} = \vec{0},$ then $\vec{r}$ is equal to

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Contributor-Level 10

$\vec{a} * [(\vec{r} - \vec{b}) * \vec{a}] + \vec{b} * [(\vec{r} - \vec{c}) * \vec{b}] + \vec{c} * [(\vec{r} - \vec{a}) * \vec{c}] = \vec{0}$

As ${| \vec{a} |}^{2} = {| \vec{b} |}^{2} = {| \vec{c} |}^{2}$

$= {| \vec{a} |}^{2} (3 \vec{r} - (\vec{a} + \vec{b} + \vec{c})) - ((\vec{a} . \vec{r}) \vec{a} + (\vec{a} . \vec{r}) \vec{b} + (\vec{c} . \vec{r}) \vec{c})$

Let $\vec{r} = \frac{\vec{a} + \vec{b} + \vec{c}}{2}$

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New answer posted

2 months ago

Let $\vec{a}, \vec{b}, \vec{c}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $θ,$ with the vector $\vec{a} + \vec{b} + \vec{c} .$ Then 36cos² 2q is equal to………….

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Raj Pandey

Contributor-Level 9

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

$\vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0$

${| \vec{a} + \vec{b} + \vec{c} |}^{2} = 3 l^{2}$

$\Rightarrow l^{2} = \sqrt[]{3} l^{2} c o s θ \Rightarrow c o s θ = \frac{1}{\sqrt[]{3}}$

36 cos² 2q = 36 ${(\frac{2}{3} - 1)}^{2} = 4$

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New answer posted

2 months ago

Let $\vec{a} = \hat{i} + 5 \hat{j} + α \hat{k}, \vec{b} = \hat{i} + 3 \hat{j} + β \hat{k} a n d \vec{c} = - \hat{i} + 2 \hat{j} - 3 \hat{k}$ be three vectors such that, $| \vec{b} * \vec{c} | = 5 \sqrt[]{3}$ and $\vec{a}$ is perpendicular to $\vec{b} .$ Then the greatest among the value of ${| \vec{a} |}^{2}$ is

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Contributor-Level 10

$\vec{b} * \vec{c} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & β \\ - 1 & 2 & - 3 \end{matrix} | = \hat{i} (- 9 - 2 β) - \hat{j} (- 3 + β) + \hat{k} (5)$

$| \vec{b} * \vec{c} | = 5 \sqrt[]{3} g i v e s \sqrt[]{{(9 - 2 β)}^{2} + {(β - 3)}^{2} + 25} = 5 \sqrt[]{3}$

$81 + 4 β^{2} + 36 β + β^{2} + 9 - 6 β + 25 = 75$

$β = - 2, - 4$

also $\vec{a} ⊥ \vec{b} s o \vec{a} . \vec{b} = 0 i . e . 1 + 15 + α β = 0$

So, ${| \vec{a} |}^{2} = 1 + 25 + α^{2} = 90$

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New answer posted

2 months ago

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k} a n d \vec{b} = \hat{j} - \hat{k} . I f \vec{c}$ is a vector such that $\vec{a} * \vec{c} = \vec{b} a n d \vec{a} . \vec{c} = 3,$ then $\vec{a} . (\vec{b} * \vec{c}) i s$ equal to

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Contributor-Level 10

$\vec{a} . \vec{b} * \vec{c}$

$= \vec{b} . \vec{c} * \vec{a}$

$= \vec{b} . (- \vec{b})$

$= - {| \vec{b} |}^{2} = 2$

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New answer posted

2 months ago