If a-2k- b - 2-k- then c-2-3k- then the angle between 2a-c and a-b is-

2 a c=2( î+ĵ+2k̂) ( 2î+ĵ+3k̂)=ĵ+k̂ a+b=( î+ĵ+2k̂)+(2î ĵ k̂)=î+k̂ If θ is the angle between 2a c, a+b then cos θ =(0)(1)+(1)(0)+(1)(1)/√(1+1)√(1+1)=1/2⇒θ =π/3 ...

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

Byju's Answer

Standard XIII

Mathematics

Angle between Two Vectors

If a=-î+ĵ+2k̂...

Question

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:

\frac{π}{4}

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

\frac{π}{3}

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

\frac{π}{2}

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

\frac{3 π}{4}

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

Open in App

Solution

The correct option is B $\frac{π}{3}$
$2 a - c = 2 (-^i +^j + 2^k) - (- 2^i +^j + 3^k) =^j +^k$
$a + b = (-^i +^j + 2^k) + (2^i -^j -^k) =^i +^k$
If $θ$ is the angle between 2a-c, a+b then $c o s θ = \frac{(0) (1) + (1) (0) + (1) (1)}{\sqrt{1 + 1} \sqrt{1 + 1}} = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$

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:

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

Find $\to a . (\to b \times \to c)$ , if $\to a = 2^i +^j + 3^k, \to b = -^i + 2^j +^k$ and $\to c = 3^i +^j + 2^k$ .

Q. If

\to r = 3^i + 2^j - 5^k, \to a = 2^i -^j +^k, \to b =^i + 3^j - 2^k

and

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

such that

\to r = x \to a + y \to b + z \to c

, then

Join BYJU'S Learning Program

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:

The correct option is B π32a−c=2(−^i+^j+2^k)−(−2^i+^j+3^k)=^j+^k a+b=(−^i+^j+2^k)+(2^i−^j−^k)=^i+^k If θ is the angle between 2a-c, a+b then cosθ=(0)(1)+(1)(0)+(1)(1)√1+1√1+1=12⇒θ=π3

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: