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.