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CBSE Sample Paper 2024-25 · Section A · 1 mark

Q9: α\alpha for an obtuse angle between p⃗\vec p and q⃗\vec q

Sample 2024-25Ch 10 · Vector Algebra
The question

The value of α\alpha if the angle between p⃗=2α2i^−3αj^+k^\vec p = 2\alpha^2\hat i - 3\alpha\hat j + \hat k and q⃗=i^+j^+αk^\vec q = \hat i + \hat j + \alpha\hat k is obtuse, is

(A) R−[0,1]R - [0, 1]

(B) (0,1)(0, 1)

(C) [0,∞)[0, \infty)

(D) [1,∞)[1, \infty)

Exam answer

For an obtuse angle, cos⁡θ<0\cos\theta < 0⁠, so p⃗⋅q⃗<0\vec p \cdot \vec q < 0⁠.

Why?

The scalar product links the angle to the components:

p⃗⋅q⃗=∣p⃗∣ ∣q⃗∣cos⁡θ\vec p \cdot \vec q = |\vec p|\,|\vec q|\cos\theta

Both lengths are positive, so p⃗⋅q⃗\vec p \cdot \vec q has the same sign as cos⁡θ\cos\theta⁠. An obtuse angle (between 90∘90^\circ and 180∘180^\circ⁠) has a negative cosine.

p⃗⋅q⃗=2α2(1)−3α(1)+1(α)=2α2−3α+α\begin{aligned} \vec p \cdot \vec q &= 2\alpha^2(1) - 3\alpha(1) \\ &\quad + 1(\alpha) \\ &= 2\alpha^2 - 3\alpha + \alpha \end{aligned}
Why?

Multiply the matching components and add: i^\hat i with i^\hat i⁠, j^\hat j with j^\hat j⁠, k^\hat k with k^\hat k⁠. The components of q⃗\vec q are 1,1,α1, 1, \alpha⁠.

2α2−3α+α<0⇒ 2α2−2α<0⇒ 2α(α−1)<0⇒ 0<α<1\begin{aligned} &2\alpha^2 - 3\alpha + \alpha < 0 \\ \Rightarrow\ &2\alpha^2 - 2\alpha < 0 \\ \Rightarrow\ &2\alpha(\alpha - 1) < 0 \\ \Rightarrow\ &0 < \alpha < 1 \end{aligned}
Why?

Collect the α\alpha terms, then take out the common factor 2α2\alpha⁠:

−3α+α=−2α2α2−2α=2α⋅α−2α⋅1=2α(α−1)\begin{aligned} -3\alpha + \alpha &= -2\alpha \\ 2\alpha^2 - 2\alpha &= 2\alpha \cdot \alpha - 2\alpha \cdot 1 \\ &= 2\alpha(\alpha - 1) \end{aligned}

A product is negative when its two factors have opposite signs. 2α>02\alpha > 0 and α−1<0\alpha - 1 < 0 gives 0<α<10 < \alpha < 1⁠. The other case, α<0\alpha < 0 and α>1\alpha > 1⁠, is impossible.

Also, p⃗\vec p is never a negative multiple of q⃗\vec q here, so the angle is never 180∘180^\circ⁠.

Watch out: option (A) is where 2α(α−1)>02\alpha(\alpha - 1) > 0⁠, the acute case.

Answer: (B) (0,1)(0, 1)

Where this comes from

Chapter 10, Vector Algebra: Scalar product: obtuse angle between vectors.

Question type: The angle between two vectors

How to spot it: click to see

Type 4. The angle between two vectors

Spot it: "find the angle between the vectors".

Here: An obtuse angle has cos⁡θ<0\cos\theta \lt 0⁠, and the lengths are positive, so the dot product 2α2−2α2\alpha^2 - 2\alpha must be negative. That gives 0<α<10 \lt \alpha \lt 1⁠.

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