Skip to content
Exercise: Dot Product · Q24

Q.For what value of λ\lambda are the vectors a⃗=2i^+λj^+k^\vec a=2\hat i+\lambda\hat j+\hat k and b⃗=i^−2j^+3k^\vec b=\hat i-2\hat j+3\hat k perpendicular?

West Bengal WbchseTextbookSubjectiveImportance★★★★★
51% · 26/51 Questions
✓ Free question

Concept understanding — Scalar (Dot) Product

For non-zero vectors a⃗,b⃗\vec a,\vec b with included angle θ\theta (0≤θ≤π0\le\theta\le\pi), the scalar (dot) product is the number a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ.\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta.

Geometric meaning (projection). a⃗⋅b⃗=∣a⃗∣×(projection of b⃗ on a⃗)\vec a\cdot\vec b=|\vec a|\times(\text{projection of }\vec b\text{ on }\vec a), and the projection of b⃗\vec b on a⃗\vec a is a⃗⋅b⃗∣a⃗∣\dfrac{\vec a\cdot\vec b}{|\vec a|} (symmetrically, projection of a⃗\vec a on b⃗\vec b is a⃗⋅b⃗∣b⃗∣\dfrac{\vec a\cdot\vec b}{|\vec b|}).

Core properties.

  • Commutative: a⃗⋅b⃗=b⃗⋅a⃗\vec a\cdot\vec b=\vec b\cdot\vec a.
  • Sign follows the angle: positive for 0≤θ<π/20\le\theta<\pi/2, zero at θ=π/2\theta=\pi/2, negative for π/2<θ≤π\pi/2<\theta\le\pi. In particular a⃗⋅b⃗=0  ⟺  a⃗=0⃗\vec a\cdot\vec b=0 \iff \vec a=\vec 0 or b⃗=0⃗\vec b=\vec 0 or a⃗⊥b⃗\vec a\perp\vec b — for two non-zero vectors, a⃗⋅b⃗=0\vec a\cdot\vec b=0 is exactly the perpendicularity test.
  • a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a=|\vec a|^2 (often written a2a^2), so ∣a⃗∣=a⃗⋅a⃗|\vec a|=\sqrt{\vec a\cdot\vec a}.
  • i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1 and i^⋅j^=j^⋅k^=k^⋅i^=0\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0 (they're mutually perpendicular unit vectors).
  • Distributive: a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c, and likewise for subtraction and for the right factor.
  • Identities (proved exactly like (x+y)2(x+y)^2 for numbers): ∣a⃗+b⃗∣2=∣a⃗∣2+∣b⃗∣2+2a⃗⋅b⃗|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b; ∣a⃗−b⃗∣2=∣a⃗∣2+∣b⃗∣2−2a⃗⋅b⃗|\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b; (a⃗+b⃗)⋅(a⃗−b⃗)=∣a⃗∣2−∣b⃗∣2(\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2.
  • Coordinate formula: for a⃗=a1i^+a2j^+a3k^\vec a=a_1\hat i+a_2\hat j+a_3\hat k, b⃗=b1i^+b2j^+b3k^\vec b=b_1\hat i+b_2\hat j+b_3\hat k: a⃗⋅b⃗=a1b1+a2b2+a3b3.\vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3.
  • Angle formula: θ=cos⁡−1 ⁣(a⃗⋅b⃗∣a⃗∣∣b⃗∣)\theta=\cos^{-1}\!\left(\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}\right).
  • Triangle/Cauchy–Schwarz-type inequalities: ∣a⃗+b⃗∣≤∣a⃗∣+∣b⃗∣|\vec a+\vec b|\le|\vec a|+|\vec b| and ∣a⃗⋅b⃗∣≤∣a⃗∣∣b⃗∣|\vec a\cdot\vec b|\le|\vec a||\vec b|.

Because the dot product pins down the angle unambiguously between 00 and π\pi, it is the preferred tool whenever a problem asks for 'the angle between two vectors' (the cross product only ever returns the acute angle, since sin⁡θ≥0\sin\theta\ge0 throughout [0,π][0,\pi]).

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.