If A-k- and B-2-3k- Find the angle between A and B-

Given, nbsp;A=î+ĵ+k̂ and nbsp;B=2î ĵ+3k̂ We know that, nbsp; cosθ=A.B|A||B| ⇒ cosθ=(î+ĵ+k̂).(2î ĵ+3k̂)√(1^2+1^2+1^2)×√(2^2+( 1)^2+3^2) ⇒ cosθ=2 1+3√((3)(1 ...

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

Byju's Answer

Standard XI

Physics

Multiplication with Vectors

If A=î+ĵ+k̂ a...

Question

If $\to A =^i +^j +^k$ and $\to B = 2^i -^j + 3^k$ . Find the angle between $\to A$ and $\to B$ .

cos - 1 (\frac{4}{\sqrt{42}})

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

cos - 1 (\frac{2}{\sqrt{14}})

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

cos - 1 (\frac{2}{\sqrt{42}})

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

cos - 1 (\frac{4}{\sqrt{3}})

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

Open in App

Solution

The correct option is A $cos - 1 (\frac{4}{\sqrt{42}})$
Given, $\to A =^i +^j +^k$ and $\to B = 2^i -^j + 3^k$
We know that,
$cos θ = \frac{\to A . \to B}{| \to A | | \to B |}$
$\Rightarrow cos θ = \frac{(^i +^j +^k) . (2^i -^j + 3^k)}{\sqrt{1^{2} + 1^{2} + 1^{2}} \times \sqrt{2^{2} + (- 1)^{2} + 3^{2}}}$
$\Rightarrow cos θ = \frac{2 - 1 + 3}{\sqrt{(3) (14)}} = \frac{4}{\sqrt{42}}$
$\Rightarrow θ = cos - 1 (\frac{4}{\sqrt{42}})$

Suggest Corrections

Similar questions

Q. If

a = -^i +^j + 2^k

b = 2^i -^j -^k

then

c = - 2^i +^j + 3^k

, then the angle between 2a-c and a+b is:

Q. If

a = -^i +^j + 2^k

b = 2^i -^j -^k

then

c = - 2^i +^j + 3^k

, then the angle between 2a-c and a+b is:

If $\to r \times \to b = \to c \times \to b$ and $\to r . \to a = 0$ where $\to a = 2^i + 3^j -^k, \to b = 3^i -^j +^k$ and $\to c =^i +^j + 3^k$ , then $\to r$ is equal to

Q. Find

\to a . (\to b \times \to c)

,if

\to a = 2^i +^j + 3^k, \to b = -^i + 2^j +^k a n d \to c = 3^i +^j + 2^k

Q. If

\to A =^i +^j +^k

and

\to B = 2^i -^j + 3^k

. Find the angle between

\to A

and

\to B

Join BYJU'S Learning Program

If →A=^i+^j+^k and →B=2^i−^j+3^k. Find the angle between →A and →B.

The correct option is A cos−1(4√42)Given, →A=^i+^j+^k and →B=2^i−^j+3^k We know that, cosθ=→A.→B|→A||→B| ⇒ cosθ=(^i+^j+^k).(2^i−^j+3^k)√12+12+12×√22+(−1)2+32 ⇒ cosθ=2−1+3√(3)(14)=4√42 ⇒ θ=cos−1(4√42)

If $\to A =^i +^j +^k$ and $\to B = 2^i -^j + 3^k$ . Find the angle between $\to A$ and $\to B$ .