If a-a1-a2-a3k-b-b1-b2-b3k- c-c1-c2-c3k- and c- is a unit vector orthogonal to the plane of a-b- and the angle between a- and b- is -3 - then -c-2 -

The correct option is C 34(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) Take Cross product #160; |a⃗×b⃗|^2=|c⃗|^2 |a⃗|^2|b⃗|^2 #160; sin^2 #160; π ...

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Byju's Answer

Standard XII

Physics

Unit Vectors

If a⃗=a1î+a...

Question

If $\to a = a_{1}^i + a_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k$ , $\to c = c_{1}^i + c_{2}^j + c_{3}^k$ and $\to c$ is a unit vector orthogonal to the plane of $\to a, \to b$ and the angle between $\to a$ and $\to b$ is $\frac{π}{3}$ , then $| \to c |^{2}$ =

0

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\frac{3}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})

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\frac{3}{4} (C_{1}^{2} a_{1}^{2} + b_{1}^{2} + X a_{2}^{2} + b_{2}^{2} + c_{2}^{2})

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Solution

The correct option is C $\frac{3}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$
Take Cross product
$| \to a \times \to b |^{2} = | \to c |^{2}$
$| \to a |^{2} | \to b |^{2} {sin}^{2} \frac{π}{3} = | \to c |^{2}$
$\frac{3}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) = | \to c |^{2}$

Suggest Corrections

Similar questions

Q. Let the vectors

\to a, \to b, \to c

given as

a_{1}^i + a_{2}^j + a_{3}^k, b_{1}^i + b_{2}^j + b_{3}^k, c_{1}^i + c_{2}^j + c_{3}^k

. Then show that

= \to a \times (\to b + \to c) = \to a \times \to b + \to a \times \to c

Q. If

\to a, \to b, \to c

are unit vectors such that

\to a + \to b + \to c = \to 0

and

(\to a, \to b) = \frac{π}{3}

, then

∣ ∣ \to a \times \to b ∣ ∣ + ∣ ∣ \to b \times \to c ∣ ∣ + | \to c \times \to a | =

Q. If

\to a, \to b, \to c

are unit vectors such that

\to a

is perpendicular to the

\to b

and

\to c

and angle between

\to b

and

\to c

\frac{π}{3}

, then value of

| \to a +

\to b +

\to c |

Q. If

\to a, \to b, \to c

are unit vectors and the angles between

\to a, \to b

and

\to b, \to c

and

\to c, \to a

are

\frac{π}{6}, \frac{π}{3}

and

\frac{π}{4}

respectively, then

| \to a + \to b + \to c |

Q. Let

\to a, \to b

and

\to c

be three vectors such that

| \to a | = \sqrt{3}, | \to b | = 5, \to b . \to c = 10

and the angle between

\to b

and

\to c

\frac{π}{3} .

\to a

is perpendicular to vector

\to b \times \to c

, then

| \to a \times (\to b \times \to c) |

is equal to

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If →a=a1^i+a2^j+a3^k,→b=b1^i+b2^j+b3^k, →c=c1^i+c2^j+c3^k and →c is a unit vector orthogonal to the plane of →a,→b and the angle between →a and →b is π3 , then |→c|2 =

The correct option is C 34(a21+a22+a23)(b21+b22+b23)Take Cross product |→a×→b|2=|→c|2|→a|2|→b|2sin2π3=|→c|234(a21+a22+a23)(b21+b22+b23)=|→c|2

If $\to a = a_{1}^i + a_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k$ , $\to c = c_{1}^i + c_{2}^j + c_{3}^k$ and $\to c$ is a unit vector orthogonal to the plane of $\to a, \to b$ and the angle between $\to a$ and $\to b$ is $\frac{π}{3}$ , then $| \to c |^{2}$ =