If -a-5-a-b -8 and -a-b -10- then -b- is equal to-

Given: nbsp;|a|=5,|a nbsp;b nbsp;|=8 and |a+b nbsp;|=10 We know, for any two vectors, nbsp;|a+b nbsp;|^2+|a b nbsp;|^2=2(|a nbsp; |^2+| b nbs ...

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Byju's Answer

Standard XII

Physics

Vector Notation

If |a|=5,|a-b...

Question

If $| \to a | = 5, | \to a - \to b | = 8$ and $| \to a + \to b | = 10$ , then $| \to b |$ is equal to:

1

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\sqrt{57}

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Solution

The correct option is D $\sqrt{57}$
Given: $| \to a | = 5, | \to a - \to b | = 8$ and $| \to a + \to b | = 10$
We know, for any two vectors, $| \to a + \to b |^{2} + | \to a - \to b |^{2} = 2 (| \to a |^{2} + | \to b |^{2}) \Rightarrow 10^{2} + 8^{2} = 2 (5^{2} + | \to b |^{2}) \Rightarrow 100 + 64 = 50 + 2 | \to b |^{2} \Rightarrow | \to b |^{2} = 57 \Rightarrow | \to b | = \sqrt{57}$

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Similar questions

Q. If

| \to a | = 5, | \to a - \to b | = 8

and

| \to a + \to b | = 10

, then

| \to b |

is equal to:

Q. If

| \to a | = 2, | \to b | = 5

and

| \to a \times \to b | = 8,

then

| \to a \cdot \to b |

is equal to

Q. If

| \to a | = | \to b | = | \to a + \to b |

=1, then

| \to a - \to b |

is equal to

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

\to a

and

\to b

are non-zero non-collinear vectors such that |

\to a

|=2,

\to a . \to b

=1 and angle between

\to a

and

\to b

\frac{π}{3}

. If

\to r

any vector satisfying

\to r

\to a = 2, \to r . \to b

= 8,

(\to r + 2 \to a - 10) . (\to a \times \to b)

= 6 and is equal to

\to r + 2 \to a - 10 \to b = | (\to a \times \to b) |

= then I=

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If |→a|=5,|→a−→b|=8 and |→a+→b|=10, then |→b| is equal to:

The correct option is D √57Given: |→a|=5,|→a−→b|=8 and |→a+→b|=10 We know, for any two vectors, |→a+→b|2+|→a−→b|2=2(|→a|2+|→b|2)⇒102+82=2(52+|→b|2)⇒100+64=50+2|→b|2⇒|→b|2=57⇒|→b|=√57

If $| \to a | = 5, | \to a - \to b | = 8$ and $| \to a + \to b | = 10$ , then $| \to b |$ is equal to: