A parallelogram is constructed on 3a-b- and a-4b- where -a-6 and -b-8 and a- and b- are anti-parallel- Then the length of the longer diagonal is-

The correct option is A 40 R⃗ =x⃗ +y⃗ #160; we have #160; x⃗ =3a⃗ +b⃗ #160; y⃗ =a⃗ 4b⃗ #160; R⃗ =3a⃗ +b⃗ +a⃗ 4b⃗ ...

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Addition of Vectors

A parallelogr...

Question

A parallelogram is constructed on $3 \to a + \to b$ and $\to a - 4 \to b$ , where $| \to a | = 6$ and $| \to b | = 8$ and $\to a$ and $\to b$ are anti-parallel. Then the length of the longer diagonal is?

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Solution

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Similar questions

5 \to a + 2 \to b

\to a - 3 \to b

| \to a | = 2 \sqrt{2}

| \to b | = 3

(\to a, \to b) = \frac{π}{4}

3 \to a - 5 \to b

2 \to a + \to b

\to a + 4 \to b

\to b - \to a

\to a

\to b

3 \to a - 5 \to b

2 \to a + \to b

\to a + 4 \to b

\to b - \to a

\to a

\to b

\to a \times \to b = \to c, \to b \times \to c = \to a

| 3 \to a + 4 \to b + 12 \to c |

\to a, \to b, \to c

\to p = 3 \to a - 5 \to b; \to q = 2 \to a + \to b; \to r = \to a + 4 \to b; \to s = - \to a + \to b

sin (\to p \land \to q) = 1

sin (\to r \land \to s) = 1

cos (\to a \land \to b)

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