# Vectors – Problems

## Vectors – Problems

Key points: let $$\overrightarrow{a}$$, $$\overrightarrow{b}$$ be two vectors dot product (or) scalar product (or) direct product (or) inner product denoted by $$\overrightarrow{a}.\overrightarrow{b}$$ which is defined as $$|\overrightarrow{a}||\overrightarrow{b}|\cos \theta$$ where $$\theta =(\overrightarrow{a}.\overrightarrow{b})$$.

(i) The product $$\overrightarrow{a}$$, $$\overrightarrow{b}$$ is zero when $$|\overrightarrow{a}|=0$$ (or) $$|\overrightarrow{b}|=0$$ (or) θ = 90°.

(ii) The angle between the vectors is $$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|.|\overrightarrow{b}|}$$.

Problems:

1. Find the angle between the vectors i + 2j + 3k and 3i – j + 2k.

Solution: Given that

i + 2j + 3k and 3i – j + 2k.

Let θ be the angle between the vectors.

Then $$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|.|\overrightarrow{b}|}$$.

$$\cos \theta =\frac{(i+2j+3k).(3i-j+2k)}{\sqrt{i+2j+3k}.\sqrt{3i-j+2k}}$$.

$$\cos \theta =\frac{|3-2+6|}{\sqrt{14}.\sqrt{14}}$$.

cosθ = 7/ 14 = ½

θ = cos (½)

θ = cos (cos⁻¹ (60°))

θ = 60°.

2. If the vectors 2i + λj – k and 4i – 2j + 2k are perpendicular to each other, then find λ.

Solution: Given that,

2i + λj – k and 4i – 2j + 2k

By hypothesis $$\overrightarrow{a}$$, $$\overrightarrow{b}$$ are perpendicular then $$\overrightarrow{a}.\overrightarrow{b}=0$$.

(2i + λj – k) (4i – 2j + 2k) = 0

8 – 2λ – 2 = 0

6 – 2λ = 0

λ = 3.