Find angle -between the vectors a - - - - - k- and b - - - - - k-

It is given that a = î + ĵ k̂ nbsp;and b = î ĵ + k̂ Magnitude of a = √(1^2+2^2+( 1)^2) = √(3) Magnitude nbsp; of b = √(1^2 + ( 1)^2 + 1^2) = √(3) Now, a ...

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Standard VII

History

Mughals

Find angle 'θ...

Question

Find angle $^{'} θ^{'}$ between the vectors $\to a =^i +^j -^k$ and $\to b =^i -^j +^k$ .

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Solution

It is given that $\to a =^i +^j -^k$ and $\to b =^i -^j +^k$
Magnitude of $\to a = \sqrt{1^{2} + 2^{2} + (- 1)^{2}} = \sqrt{3}$
Magnitude of
$\to b = \sqrt{1^{2} + (- 1)^{2} + 1^{2}} = \sqrt{3}$
Now,
$\to a . \to b = (^i +^j -^k) . (^i -^j -^k)$
$\to a . \to b = 1.1 + 1. (- 1) + (- 1) 1$
$\to a . \to b = - 1$
The angle between two vectors is
$cos θ = \frac{\to a . \to b}{∣ ∣ \begin{matrix} \to a \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} \to b \end{matrix} ∣ ∣}$
$cos θ = \frac{- 1}{\sqrt{3} . \sqrt{3}}$
$cos θ = \frac{1}{3}$
$θ = {cos}^{- 1} (\frac{- 1}{3})$

Final answer:
Hence, the angle between the two vectors is ${cos}^{- 1} (\frac{- 1}{3})$

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Mughals

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Find angle ′θ′between the vectors →a=^i+^j−^k and →b=^i−^j+^k.

Find angle $^{'} θ^{'}$ between the vectors $\to a =^i +^j -^k$ and $\to b =^i -^j +^k$ .