If -A-B-A-B- then -A-B- is-

The correct option is C √(|A⃗| ^2+|B⃗| ^2+√(2) |A⃗||B⃗|) Given: | A⃗×B⃗ #160;| =A⃗·B⃗ or ABsinθ =ABcosθ #160; #160; #160;or tanθ #16 ...

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Byju's Answer

Standard XII

Mathematics

Determinant

If |A⃗×B⃗|=...

Question

If $| \to A \times \to B | = \to A . \to B$ , then $| \to A + \to B |$ is:

| \to A | + | \to B |

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\sqrt{| \to A |^{2} + | \to B |^{2}}

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\sqrt{| \to A |^{2} + | \to B |^{2} + \frac{| \to A | | \to B |}{\sqrt{2}}}

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\sqrt{| \to A |^{2} + | \to B |^{2} + \sqrt{2} | \to A | | \to B |}

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Solution

The correct option is C $\sqrt{| \to A |^{2} + | \to B |^{2} + \sqrt{2} | \to A | | \to B |}$
Given: $∣ ∣ \to A \times \to B ∣ ∣ = \to A \cdot \to B o r A B sin θ = A B cos θ o r tan θ = 1 o r θ = 45^{o}$
Now, $∣ ∣ \to A + \to B ∣ ∣ = \sqrt{{∣ ∣ \to A ∣ ∣}^{2} + {∣ ∣ \to B ∣ ∣}^{2} + 2 ∣ ∣ \to A ∣ ∣ ∣ ∣ \to B ∣ ∣ cos θ} = \sqrt{{∣ ∣ \to A ∣ ∣}^{2} + {∣ ∣ \to B ∣ ∣}^{2} + \sqrt{2} ∣ ∣ \to A ∣ ∣ ∣ ∣ \to B ∣ ∣}$

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Similar questions

Q. I : If

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

, then

(\to a, \to b) = \frac{π}{2}

.
II : If

\to a, \to b, \to a + \to b

are unit vectors, then

(\to a, \to b) = \frac{2 π}{3}

Q. For vectors

\to a

\to b

. Prove that

| \to a \times \to b |^{2} = | \to a |^{2} | \to b |^{2} - | \to a . \to b |^{2}

The vectors $\to a$ and $\to b$ have magnitudes 2 and 2 $\sqrt{2}$ respectively. It is found that $\to a$ . $\to b$ = | $\to a$ $\times$ $\to b$ |. Then the value of | $\frac{\to a + \to b}{\to a - \to b}$ | will be

Q. Let

\to a

and

\to b

be two unit vectors. the maximum value of

\frac{{∣ ∣ \to a + \to b ∣ ∣}^{2} - {∣ ∣ \to a - \to b ∣ ∣}^{2}}{{∣ ∣ \to a + \to b ∣ ∣}^{2} + {∣ ∣ \to a - \to b ∣ ∣}^{2}}

is equal to

Q. Let

| \to a | = 1, ∣ ∣ \to b ∣ ∣ = \sqrt{2}

| \to c | = \sqrt{3}

, and

\to a ⊥ (\to b + \to c)

\to b ⊥ (\to c + \to a)

and

\to c ⊥ (\to a + \to b)

, then

∣ ∣ \to a + \to b + \to c ∣ ∣

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If |→A×→B|=→A.→B, then |→A+→B| is:

If $| \to A \times \to B | = \to A . \to B$ , then $| \to A + \to B |$ is: