Let a- and b- be two vectors such that -a-3-b-6 and -a-b-7- Then the value of 3 a-2 b- -2 a-5 b-4 a-b- is

The correct option is C 284(3a 2b) · (2a+5b 4a×b) =6|a|^2+15a·b 3a· (4a×b) 4a·b 10|b|^2+2b· (4a×b) Now 3a· (4a×b)=0 and nbsp;2b· (4a×b)=0 because n ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

Let a⃗ and b⃗...

Question

Let $\to a$ and $\to b$ be two vectors such that $| \to a | = 3$ , $| \to b | = 6$ and $| \to a + \to b | = 7$ . Then the value of $(3 \to a - 2 \to b) . (2 \to a + 5 \to b - 4 \to a \times \to b)$ is

- 306

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

- 295

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

- 284

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

- 290

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

Open in App

Solution

The correct option is C $- 284$
$(3 \to a - 2 \to b) \cdot (2 \to a + 5 \to b - 4 \to a \times \to b)$
$= 6 | \to a |^{2} + 15 \to a \cdot \to b - 3 \to a \cdot (4 \to a \times \to b) - 4 \to a \cdot \to b - 10 | \to b |^{2} + 2 \to b \cdot (4 \to a \times \to b)$

Now $3 \to a \cdot (4 \to a \times \to b) = 0$ and $2 \to b \cdot (4 \to a \times \to b) = 0$ because $\to a$ and $\to b$ will be perpendicular to $4 \to a \times \to b$

So the equation reduces to
$(3 \to a - 2 \to b) \cdot (2 \to a + 5 \to b - 4 \to a \times \to b) = 6 | \to a |^{2} + 15 \to a \cdot \to b - 4 \to a \cdot \to b - 10 | \to b |^{2}$
$= 6 (3)^{2} + 11 \to a \cdot \to b - 10 (6)^{2}$
$= 11 \to a \cdot \to b - 306 \dots (1)$

Now $| \to a + \to b | = 7$
$\Rightarrow | \to a + \to b |^{2} = 7^{2}$
$\Rightarrow | \to a |^{2} + 2 \to a \cdot \to b + | \to b |^{2} = 7^{2}$
$\Rightarrow 3^{2} + 2 \to a \cdot \to b + 6^{2} = 7^{2}$
$\Rightarrow \to a \cdot \to b = 2$
Putting this value in equation $(1),$ we get
$(3 \to a - 2 \to b) \cdot (2 \to a + 5 \to b - 4 \to a \times \to b) = 11 \times 2 - 306 = - 284$

Suggest Corrections

Similar questions

Q. If

\to a

and

\to b

are two non collinear unit vectors and

| \to a + \to b | = \sqrt{3}

, then

(2 \to a - 5 \to b) . (3 \to a + \to b) =

Q. Let

\to a

and

\to b

be unit vectors such that

| \to a + \to b | = \sqrt{3}

. Then the value of

(2 \to a + 5 \to b) \cdot (3 \to a + \to b + \to a \times \to b)

Q. If

\to a

and

\to b

are unit vectors such that

(\to a + \to b) . (2 \to a + 3 \to b) \times (3 \to a - 2 \to b) = 0

Q. Let

\to a

and

\to b

be two unit vectors such that

| \to a + \to b | = \sqrt{3} .

\to c = \to a + 2 \to b + 3 (\to a \times \to b),

then

2 | \to c |

is equal to:

Q. If

\to a, \to b, \to c

are non-coplanar vectors, then show that the four points

2 \to a + \to b, \to a + 2 \to b + \to c, 4 \to a - 2 \to b - \to c

and

3 \to a + 4 \to b - 5 \to c

are coplanar.

Join BYJU'S Learning Program

Let →a and →b be two vectors such that |→a|=3, |→b|=6 and |→a+→b|=7. Then the value of (3→a−2→b).(2→a+5→b−4→a×→b) is

Let $\to a$ and $\to b$ be two vectors such that $| \to a | = 3$ , $| \to b | = 6$ and $| \to a + \to b | = 7$ . Then the value of $(3 \to a - 2 \to b) . (2 \to a + 5 \to b - 4 \to a \times \to b)$ is