CBSE Board Examination 2024 · Section A · 1 mark
Q7: a ^ ⋅ b ^ \hat a \cdot \hat b a ^ ⋅ b ^ sin θ = 3 5 \sin\theta = \frac{3}{5} sin θ = 5 3
With chapter links Answers only
The question
Let θ \theta θ a ^ \hat a a ^ b ^ \hat b b ^ sin θ = 3 5 \sin\theta = \frac{3}{5} sin θ = 5 3 a ^ ⋅ b ^ \hat a \cdot \hat b a ^ ⋅ b ^
(A) ± 3 5 \pm\frac{3}{5} ± 5 3
(B) ± 3 4 \pm\frac{3}{4} ± 4 3
(C) ± 4 5 \pm\frac{4}{5} ± 5 4
(D) ± 4 3 \pm\frac{4}{3} ± 3 4
Exam answer Show all explanations
Given sin θ = 3 5 \sin\theta = \frac{3}{5} sin θ = 5 3
cos 2 θ = 1 − sin 2 θ = 1 − 9 25 = 16 25 ⇒ cos θ = ± 4 5 \begin{aligned} \cos^2\theta &= 1 - \sin^2\theta \\ &= 1 - \frac{9}{25} = \frac{16}{25} \\ \Rightarrow \cos\theta &= \pm\frac{4}{5} \end{aligned} cos 2 θ ⇒ cos θ = 1 − sin 2 θ = 1 − 25 9 = 25 16 = ± 5 4 Why?
This uses the identity
sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1 with sin 2 θ = 9 / 25 \sin^2\theta = 9/25 sin 2 θ = 9/25
sin 2 θ = ( 3 5 ) 2 = 3 2 5 2 = 9 25 cos 2 θ = 1 − 9 25 = 25 25 − 9 25 = 25 − 9 25 = 16 25 cos θ = ± 16 25 = ± 16 25 = ± 4 5 \begin{aligned} \sin^2\theta &= \Big(\frac35\Big)^2 = \frac{3^2}{5^2} = \frac{9}{25} \\[4pt] \cos^2\theta &= 1 - \frac{9}{25} \\ &= \frac{25}{25} - \frac{9}{25} \\ &= \frac{25 - 9}{25} = \frac{16}{25} \\[4pt] \cos\theta &= \pm\sqrt{\frac{16}{25}} \\ &= \pm\frac{\sqrt{16}}{\sqrt{25}} = \pm\frac{4}{5} \end{aligned} sin 2 θ cos 2 θ cos θ = ( 5 3 ) 2 = 5 2 3 2 = 25 9 = 1 − 25 9 = 25 25 − 25 9 = 25 25 − 9 = 25 16 = ± 25 16 = ± 25 16 = ± 5 4 Both signs stay. The angle between two vectors lies in [ 0 , π ] [0, \pi] [ 0 , π ] sin θ = 3 / 5 \sin\theta = 3/5 sin θ = 3/5 cos θ = 4 / 5 \cos\theta = 4/5 cos θ = 4/5 cos θ = − 4 / 5 \cos\theta = -4/5 cos θ = − 4/5
Since ∣ a ^ ∣ = ∣ b ^ ∣ = 1 |\hat a| = |\hat b| = 1 ∣ a ^ ∣ = ∣ b ^ ∣ = 1
a ^ ⋅ b ^ = ∣ a ^ ∣ ∣ b ^ ∣ cos θ = cos θ \hat a \cdot \hat b = |\hat a||\hat b|\cos\theta = \cos\theta a ^ ⋅ b ^ = ∣ a ^ ∣∣ b ^ ∣ cos θ = cos θ Why?
The scalar product is defined by
a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ \vec a \cdot \vec b = |\vec a||\vec b|\cos\theta a ⋅ b = ∣ a ∣∣ b ∣ cos θ A unit vector has magnitude 1 1 1 a ^ ⋅ b ^ = cos θ = ± 4 / 5 \hat a \cdot \hat b = \cos\theta = \pm 4/5 a ^ ⋅ b ^ = cos θ = ± 4/5
Watch out: option (A) uses the value of sin θ \sin\theta sin θ cos θ \cos\theta cos θ tan θ = ± 3 / 4 \tan\theta = \pm 3/4 tan θ = ± 3/4
Answer: (C) ± 4 5 \pm\dfrac{4}{5} ± 5 4
← Previous in this paper Q6Next in this paper → Q8