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 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

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?

40

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

64

No worries! We‘ve got your back. Try BYJU‘S free classes today!

32

No worries! We‘ve got your back. Try BYJU‘S free classes today!

48

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

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)

Join BYJU'S Learning Program