Q.Find angle θ between the vectors a=i^+j^−k^ and b=i^−j^+k^.
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🎯 Appeared in past exams:GUJCET 2024· Set 13· 1mreworded
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Concept understanding — Dot Product Angle
Finding the Angle Between Vectors
Suppose you have two arrows drawn from the same point. One question is unavoidable in geometry, physics, and mechanics: what is the angle between them? You could measure it with a protractor on paper, but that fails the moment the vectors live in 3D. The dot product gives you the angle by pure calculation.
The Core Idea
The scalar (dot) product of two vectors has two faces that describe the same number:
a⋅b=a1b1+a2b2+a3b3(components)
a⋅b=∣a∣∣b∣cosθ(geometry)
The first is easy to compute from coordinates; the second hides the angle θ (with 0≤θ≤π) between the vectors. Setting them equal and solving for cosθ gives the master formula.
cosθ=∣a∣∣b∣a⋅b,θ=cos−1(∣a∣∣b∣a⋅b)
Why It Works
Both vectors have a fixed length, so the only thing the dot product can "vary" with is how aligned they are. When they point the same way, cosθ=1 and the dot product is as large as possible, ∣a∣∣b∣. When they are perpendicular, cosθ=0 and the dot product vanishes. When they point opposite ways, cosθ=−1. Dividing a⋅b by the two lengths simply strips away the size information and leaves behind a pure measure of alignment — exactly cosθ.
Note
The sign of the dot product tells you the type of angle at a glance: positive ⇒ acute, zero ⇒ right angle, negative ⇒ obtuse.
Using the Formula
For a=i^+2j^+2k^ and b=i^+0j^+0k^:
a⋅b=1,∣a∣=3,∣b∣=1
cosθ=3⋅11=31⇒θ=cos−131≈70.5∘
Watch out
Never forget to divide by both magnitudes. A common slip is to compute a⋅b and call it cosθ — that is only valid if both vectors are already unit vectors.
Why You'll Use This
This single formula powers a huge range of problems: checking perpendicularity, finding the angle a line makes with an axis, computing the work done by a force at an angle, and testing whether a triangle is right-angled. Whenever the words "angle between" appear, reach for cosθ=∣a∣∣b∣a⋅b.
Finding the angle between two vectors using the dot product is one of the most exam-heavy applications in the NCERT Class 12 Vector Algebra chapter, tested in nearly every CBSE board paper and JEE Main sitting. Students searching "angle between two vectors formula and examples" should pair this with the perpendicularity and parallelism tests for a complete revision of the chapter's core toolkit.
Concept: Dot Product Angle — the cosine of the angle between two vectors is given by cosθ=∣a∣∣b∣a⋅b.
First, compute the dot product:
a⋅b=(1)(1)+(1)(−1)+(−1)(1)=1−1−1=−1.
Next, find the magnitudes:
∣a∣=12+12+(−1)2=3,∣b∣=12+(−1)2+12=3.
Then,
cosθ=3⋅3−1=−31.
Thus,
θ=cos−1(−31).
✓Final answer
The angle is θ=cos−1(−31).
The angle between two vectors is found using the dot product formula a⋅b=∣a∣∣b∣cosθ. For a=i^+j^−k^ and b=i^−j^+k^, the dot product is −1, each magnitude is 3, so cosθ=−31 and θ=cos−1(−31).
The dot product gives us a direct link between two vectors and the angle between them. When you take a⋅b, you're essentially multiplying the magnitude of one vector by the projection of the other onto it. That projection depends on cosθ, so if we know the dot product and the magnitudes, we can solve for the angle.
a⋅b=∣a∣∣b∣cosθ
This is the central relationship. Rearranging gives cosθ=∣a∣∣b∣a⋅b, and then θ=cos−1(that value).
Let's work through it.
Compute the dot product a⋅b.
For vectors in component form, multiply corresponding components and add:
a⋅b=(1)(1)+(1)(−1)+(−1)(1)=1−1−1=−1
Find the magnitude of each vector.
For a=i^+j^−k^:
∣a∣=12+12+(−1)2=1+1+1=3
For b=i^−j^+k^:
∣b∣=12+(−1)2+12=1+1+1=3
Plug into the formula.
cosθ=3⋅3−1=3−1
Write the angle.
Since cosθ=−31, we have:
θ=cos−1(−31)
Watch out
A common mistake is to forget the negative sign in the dot product. Here, a⋅b=−1, not +1. That negative tells you the angle is obtuse (greater than 90∘), which makes sense because the vectors point in somewhat opposite directions.
Tip
Notice both vectors have the same magnitude 3. When magnitudes are equal, the cosine formula simplifies to cosθ=∣a∣2a⋅b, which can save a step.
✓Final answer
The angle between the vectors is θ=cos−1(−31).
Method: Angle between two vectors given in component form
Use this to find the angle between two vectors written as i^,j^,k^ combinations.
Steps
Step 1: Compute the dot product from components.
a⋅b=a1b1+a2b2+a3b3.
Track every sign — a single sign slip flips an obtuse angle to acute.
Step 2: Compute both magnitudes.
∣a∣=a12+a22+a32,∣b∣=b12+b22+b32.
Step 3: Apply the cosine formula and invert.
cosθ=∣a∣∣b∣a⋅b,θ=cos−1(∣a∣∣b∣a⋅b).
If cosθ is not a standard value, leaving the answer as cos−1(⋅) is the correct exact form. A negative cosθ means an obtuse angle.
Common Mistakes
Mistake 1: Sign error in the dot product.
Why it's wrong: (1)(1)+(1)(−1)+(−1)(1)=1−1−1=−1, not +1; a dropped sign changes the angle from obtuse to acute. Correct approach: multiply corresponding components with their signs and sum carefully.
Mistake 2: Trying to force a "nice" degree answer.
Why it's wrong: here cosθ=−31 is not a standard cosine, so the exact angle is cos−1(−31). Correct approach: leave the answer as an inverse cosine; the negative value correctly signals an obtuse angle.