Vector Algebra

Let the vectors a,b,c be such that |a|=2, |b|=4 and |c|=4. If the projection of b on a is equal to the projection of c on a and b is perpendicular to c, then the value of |a+b-c| is

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Contributor-Level 10

6.00
b·a = c·a
|a+b-c|² = |a|²+|b|²+|c|²+2(a·b - b·c - a·c)
= 4+16+16+2(a·b - 0 - a·b) = 36
⇒ |a+b-c| = 6

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a month ago

If (a + 3b) is perpendicular to (7a - 5b) and (a - 4b) is perpendicular to (7a - 2b), then the angle between a and b (in degrees) is…….

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Contributor-Level 10

(a+3b). (7a-5b) = 7|a|² - 5ab + 21ab - 15|b|² = 7|a|²+16ab-15|b|²=0.
(a-4b). (7a-2b) = 7|a|² - 2ab - 28ab + 8|b|² = 7|a|²-30ab+8|b|²=0.
Subtracting: 46ab - 23|b|² = 0 ⇒ 2ab = |b|².
Substituting: 7|a|² + 8|b|² - 15|b|² = 0 ⇒ 7|a|² = 7|b|² ⇒ |a|=|b|.
cosθ = ab/ (|a|b|) = ab/|b|² = (1/2)|b|²/|b|² = 1/2.
θ = 60°.

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a month ago

Let a = i + j + k, b and c = j - k be three vectors such that a * b = c and a . b = 1. If the length of projection vector of the vector b on the vector a * c is l, then the value of 3l² is equal to ________.

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Contributor-Level 10

a*b=c ⇒ a.c=0, b.c=0.
|c|² = |a|²|b|² - (a.b)² = (3)|b|² - 1. |c|=√2. So |b|²=1, |b|=1.
Projection of b on a*c.
a*c = a* (a*b) = (a.b)a - (a.a)b = a - 3b.
|a-3b|² = |a|²+9|b|²-6 (a.b) = 3+9-6 = 6.
l = |b. (a-3b)|/|a-3b| = | (a.b)-3|b|²|/√6 = |1-3|/√6 = 2/√6.
3l² = 3 (4/6) = 2.

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a month ago

If |a| = 2, |b| = 5 and |a * b| = 8, then a . b is equal to:

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Contributor-Level 10

|a * b|² + |a . b|² = |a|²|b|²
8² + (a . b)² = 2² * 5²
64 + (a . b)² = 100
(a . b)² = 36
a . b = 6 (since angle seems acute from options, but could be -6).

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a month ago

Let a = i + j + 2k and b = -i + 2j + 3k. Then the vector product (a + b) * ((a * ((a - b) * b)) * b) is equal to:

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Contributor-Level 10

a = i + j + 2k
b = -i + 2j + 3k
a + b = 3j + 5k
a . b = -1 + 2 + 6 = 7
a * b = |i, j, k; 1, 2; -1, 2, 3| = -i - 5j + 3k
(a - b) * b) = (a * b) - (b * b) = a * b
(a * (a - b) * b) = a * (a * b) = (a . b)a - (a . a)b = 7a - 6b
. The expression becomes (a + b) * (7a - 6b) * b)
= (a + b) * (7 (a * b)
= 7 [ (a * (a * b) + (b * (a * b) ]
= 7 [ (

...more

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a month ago

If vectors ${\vec{a}}_{1} = x \hat{i} - \hat{j} + \hat{k} a n d {\vec{a}}_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is:

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Contributor-Level 10

${\vec{a}}_{1} = x \hat{i} - \hat{j} + \hat{k} & {\vec{a}}_{2} = \hat{i} + y \hat{j} + z \hat{k}$

given ${\vec{a}}_{1} & {\vec{a}}_{2}$ are collinear then ${\vec{a}}_{1} = λ {\vec{a}}_{2}$

$\Rightarrow (x \hat{i} - \hat{j} + \hat{k}) = λ (\hat{i} + y \hat{j} + z \hat{k})$

Since $\hat{i}, \hat{j} & \hat{k}$ are not collinear so

$S o x \hat{i} + y \hat{j} + z \hat{k} = λ \hat{i} - \frac{1}{λ} \hat{j} + \frac{1}{λ} \hat{k}$

Hence possible unit vector parallel to it be $\frac{1}{\sqrt[]{3}} (\hat{i} - \hat{j} + \hat{k})$ for $λ$ =

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New answer posted

a month ago

If $\vec{a} a n d \vec{b}$ are perpendicular, then $\vec{a} * (\vec{a} * (\vec{a} * (\vec{a} * \vec{b})))$ is equal to

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Payal Gupta

Contributor-Level 10

$\vec{a} * (\vec{a} * \vec{b}) = (\vec{a} . \vec{b}) \vec{a} - {| \vec{a} |}^{2} \vec{b} = 0 - {| \vec{a} |}^{2} \vec{b} (a s \vec{a} . \vec{b} = 0 g i v e n)$

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

$\vec{a} * (\vec{a} * (\vec{a} * (\vec{a} * \vec{b}))) = - {| \vec{a} |}^{2} \vec{a} * (\vec{a} * \vec{b}) = - {| \vec{a} |}^{2} (- {| \vec{a} |}^{2} \vec{b}) = {| \vec{a} |}^{4} \vec{b}$

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New answer posted

a month ago

Let three vectors $\vec{a}, \vec{b} a n d \vec{c}$ be such that $\vec{c}$ is coplanar with $\vec{a} a n d \vec{b}, \vec{a} . \vec{c} = 7 a n d \vec{b}$ is perpendicular to $\vec{c}, w h e r e \vec{a} = - \hat{i} + \hat{j} + \hat{k} a n d \hat{b} = 2 \hat{i} + \hat{k},$ then the value of 2 ${| \vec{a} + \vec{b} + \vec{c} |}^{2}$ is____________.

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$\vec{c} = α \vec{a} + β \vec{b} . . . . . (i)$

$\vec{a} . \vec{c} = 7 \vec{b} . \vec{c} = 0$

$\vec{a} = - \hat{i} + \hat{j} + \hat{k} \Rightarrow | \vec{a} | = \sqrt[]{3}$

$\vec{b} = 2 \hat{i} + \hat{k} \Rightarrow | \vec{b} | = \sqrt[]{5} \vec{a} . \vec{b} = - 2 + 1 = - 1$

From $(i) \vec{a} . \vec{c} = α {| \vec{a} |}^{2} - β$

$3 α - β = 7 . . . . . . . . . . (i i)$

$\bar{b} . \bar{c} = α \bar{b} . \bar{a} + β {| \vec{b} |}^{2} \Rightarrow - α + 5 β = 0 . . . . . . . (i i i)$

Solving $α = \frac{5}{2} a n d β = \frac{1}{2}$

$\Rightarrow 2 {| \vec{a} + \vec{b} + \vec{c} |}^{2} = 75$

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a month ago

Let $\vec{a} = \hat{i} + α \hat{j} + 3 \hat{k} a n d \vec{b} = 3 \hat{i} - α \hat{j} + \hat{k} .$ It the area of the parallelogram shoes adjacent sides are represented by the vectors $\vec{a} a n d \vec{b} i s 8 \sqrt[]{3}$ square units, then $\vec{a} . \vec{b}$ is equal to……….

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Payal Gupta

Contributor-Level 10

$\vec{a} * \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & α & 3 \\ 3 & - α & 1 \end{matrix} | = (4 α \hat{i} + 8 \hat{j} - 4 α \hat{k})$

$| \vec{a} * \vec{b} | = \sqrt[]{32 α^{2} + 64} = 8 \sqrt[]{3}$

322 + 64 = 192

2 = $\frac{128}{32} = 4$

$\vec{a} . \vec{b} = 3 - α^{2} + 3 = 6 - α^{2} = 6 - 4 = 2$

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a month ago

Let $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j} a n d \vec{c} = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} * \vec{a} = \vec{c} * \vec{a} a n d \vec{r} . \vec{b} = 0,$ then $\vec{r} . \vec{a}$ is equal to………………

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