Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Let → a = 2 → I + 3 → J + 4 → K and → B = 3 → I + 4 → J + 5 → K Find the Angle Between Them. - Physics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Answer in Brief

Let $\vec{a} = 2 \vec{i} + 3 \vec{j} + 4 \vec{k} \text { and } \vec{b} = 3 \vec{i} + 4 \vec{j} + 5 \vec{k}$ Find the angle between them.

Advertisement Remove all ads

#### Solution

We have:

$\vec{a} = 2 \vec{i} + 3 \vec{j} + 4 \vec{k}$

$\vec{b} = 3 \vec{i} + 4 \vec{j} + 5 \vec{k}$

Using scalar product, we can find the angle between vectors $\vec{a}$ and $\vec{b}$.

i.e.,

$\vec{a} . \vec{b} = \left| \vec{a} \right|\left| \vec{b} \right| \cos \theta$

So, $\theta = \cos^{- 1} \left( \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right|\left| \vec{b} \right|} \right)$

$= \cos^{- 1} \left( \frac{2 \times 3 + 3 \times 4 + 4 \times 5}{\sqrt{\left( 2^2 + 3^2 + 4^2 \right)} \sqrt{\left( 3^2 + 4^2 + 5^2 \right)}} \right)$

$= \cos^{- 1} \left( \frac{38}{\sqrt{29} \sqrt{50}} = \cos^{- 1} \frac{38}{\sqrt{1450}} \right)$

∴ The required angle is $\cos^{- 1} \frac{38}{\sqrt{1450}} .$

Concept: What is Physics?
Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 2 Physics and Mathematics
Exercise | Q 13 | Page 29
Share
Notifications

View all notifications

Forgot password?