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JEE Mains Maths · Vector Algebra

Solving Vector Equations

Finding an unknown vector from cross-product and dot-product conditions: r × a = b × a type equations, and a × c = b with a · c given.

All Vector Algebra notesJEE Mains Maths strategy

Why this matters

Forty-one PYQs, the largest page in the chapter. They look different but reduce to two moves: a cross product equal to zero means two vectors are parallel, and crossing a cross-product equation with a known vector turns it into one you can solve. Two ideas cover the page.

Concept 1 of 2: r × a = b × a: the difference is parallel to a

Move everything to one side: r⃗×a⃗−b⃗×a⃗=(r⃗−b⃗)×a⃗=0⃗\vec r\times\vec a-\vec b\times\vec a=(\vec r-\vec b)\times\vec a=\vec 0. A cross product is zero only when the two vectors are parallel, so r⃗=b⃗+λa⃗\vec r=\vec b+\lambda\vec a. The extra condition, usually a dot product, fixes λ\lambda. The same move handles 2(a⃗×c⃗)+3(b⃗×c⃗)=0⃗2(\vec a\times\vec c)+3(\vec b\times\vec c)=\vec0, which says c⃗∥2a⃗+3b⃗\vec c\parallel2\vec a+3\vec b.

Definition

  • u⃗×v⃗=0⃗\vec u\times\vec v=\vec0 (both non-zero) means u⃗∥v⃗\vec u\parallel\vec v.
  • r⃗×a⃗=b⃗×a⃗⇒r⃗=b⃗+λa⃗\vec r\times\vec a=\vec b\times\vec a\Rightarrow\vec r=\vec b+\lambda\vec a.
  • a⃗×c⃗=c⃗×b⃗⇒(a⃗+b⃗)×c⃗=0⃗\vec a\times\vec c=\vec c\times\vec b\Rightarrow(\vec a+\vec b)\times\vec c=\vec0.
  • Then r⃗⋅d⃗=k\vec r\cdot\vec d=k gives λ=k−b⃗⋅d⃗a⃗⋅d⃗\lambda=\frac{k-\vec b\cdot\vec d}{\vec a\cdot\vec d}.

The parallel form

r⃗×a⃗=b⃗×a⃗ ⇒ r⃗=b⃗+λa⃗\vec r\times\vec a=\vec b\times\vec a\ \Rightarrow\ \vec r=\vec b+\lambda\vec a

Worked example

r⃗×(1,1,0)=(0,0,1)×(1,1,0)\vec r\times(1,1,0)=(0,0,1)\times(1,1,0) and r⃗⋅i^=3\vec r\cdot\hat i=3. Find r⃗\vec r.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q156Moderate

Example 1 · Vector Algebra · Solving Vector Equations

If a→=ı^+2k^,b→=ı^+ȷ^+k^,c→=7ı^−3ȷ^+4k^,r→×b→+b→×c→=0→\overrightarrow{a}=\widehat{\imath}+ 2\widehat{k},\overrightarrow{b}=\widehat{\imath}+\widehat{\jmath}+\widehat{k},\overrightarrow{c}= 7\widehat{\imath}- 3\widehat{\jmath}+ 4\widehat{k},\overrightarrow{r}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}=\overrightarrow{0} and r→⋅a→=0\overrightarrow{r}\cdot\overrightarrow{a}= 0. Then r→⋅c→\overrightarrow{r}\cdot\overrightarrow{c} is equal to

c × b is minus b × c

a⃗×c⃗=c⃗×b⃗\vec a\times\vec c=\vec c\times\vec b gives (a⃗+b⃗)×c⃗=0⃗(\vec a+\vec b)\times\vec c=\vec0, not (a⃗−b⃗)(\vec a-\vec b): moving c⃗×b⃗\vec c\times\vec b across flips it to b⃗×c⃗\vec b\times\vec c.
Drill 23 more on r × a = b × a: the difference is parallel to a

Concept 2 of 2: a × c = b with a · c given: cross again with a

A cross-product equation alone does not fix c⃗\vec c: adding any multiple of a⃗\vec a leaves a⃗×c⃗\vec a\times\vec c unchanged. The dot condition a⃗⋅c⃗=k\vec a\cdot\vec c=k removes that freedom. Cross both sides with a⃗\vec a: a⃗×(a⃗×c⃗)=(a⃗⋅c⃗)a⃗−∣a⃗∣2c⃗=a⃗×b⃗\vec a\times(\vec a\times\vec c)=(\vec a\cdot\vec c)\vec a-|\vec a|^2\vec c=\vec a\times\vec b, so c⃗=ka⃗−a⃗×b⃗∣a⃗∣2\vec c=\frac{k\vec a-\vec a\times\vec b}{|\vec a|^2}. A solution exists only if b⃗⊥a⃗\vec b\perp\vec a.

Definition

  • a⃗×(a⃗×c⃗)=(a⃗⋅c⃗)a⃗−∣a⃗∣2c⃗\vec a\times(\vec a\times\vec c)=(\vec a\cdot\vec c)\vec a-|\vec a|^2\vec c.
  • a⃗×c⃗=b⃗\vec a\times\vec c=\vec b, a⃗⋅c⃗=k⇒c⃗=ka⃗−a⃗×b⃗∣a⃗∣2\vec a\cdot\vec c=k\Rightarrow\vec c=\frac{k\vec a-\vec a\times\vec b}{|\vec a|^2}.
  • Solvable only if a⃗⋅b⃗=0\vec a\cdot\vec b=0; and then c⃗⋅b⃗=0\vec c\cdot\vec b=0 too.

Solving a × c = b

c⃗=(a⃗⋅c⃗) a⃗−a⃗×b⃗∣a⃗∣2\vec c=\frac{(\vec a\cdot\vec c)\,\vec a-\vec a\times\vec b}{|\vec a|^2}

Worked example

a⃗=(1,0,1)\vec a=(1,0,1), a⃗×c⃗=(1,0,−1)\vec a\times\vec c=(1,0,-1) and a⃗⋅c⃗=2\vec a\cdot\vec c=2. Find c⃗\vec c.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 18 · Q72Moderate

Example 2 · Vector Algebra · Solving Vector Equations

Let a→=i^+j^+k^\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k} and b→=j^−k^\overrightarrow{b}=\widehat{j}-\widehat{k}. If c→\overrightarrow{c} is a vector such that a→×c→=b→\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b} and a→.c→=3\overrightarrow{a}.\overrightarrow{c}= 3, then a→⋅(b→×c→)\overrightarrow{a}\cdot (\overrightarrow{b}\times\overrightarrow{c}) is equal to:

Check solvability first

If a⃗⋅b⃗≠0\vec a\cdot\vec b\ne0, no c⃗\vec c satisfies a⃗×c⃗=b⃗\vec a\times\vec c=\vec b. A question asking how many such vectors exist can have the answer 0.
Drill 16 more on a × c = b with a · c given: cross again with a

Summary — formulas & gotchas at a glance

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Formulas (2)

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