Q.If x+3−6b−3z+4a−1−212y−700=0−62b+46−3−213y−22c+20 Find the values of a,b,c,x,y and z.
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Concept understanding — Vector Equality
Vector Equality
When are two vectors the same vector? A vector carries only two pieces of information — magnitude (length) and direction. It does not carry a fixed starting point. So two vectors are equal when they agree on both magnitude and direction, no matter where each one happens to be drawn.
Precise definition
Two vectors a and b are equal, written a=b, if and only if
they have the same magnitude, ∣a∣=∣b∣, and
they have the same direction.
Important
Position is irrelevant. An arrow drawn at the top-left of the page and an identical-length, identical-direction arrow at the bottom-right are the same vector. This is why such vectors are called "free" vectors — you may slide them anywhere without changing them.
In component form
Equality becomes a simple check on coordinates. If
a=a1i^+a2j^+a3k^,b=b1i^+b2j^+b3k^,
then
a=b⟺a1=b1,a2=b2,a3=b3.
Equal vectors have equal corresponding components — a single vector equation is really three scalar equations at once.
Don't confuse it with neighbouring ideas
Watch out
Equal vectors are not the same as coinitial vectors (which merely share a starting point) or collinear vectors (which are parallel but may differ in length or point oppositely). Two vectors of equal length pointing in opposite directions are not equal — they are negatives of each other, a=−b.
Why it matters
Because equal vectors can be moved freely, we are allowed to shift vectors to a common tail before adding them, to compare forces acting at different points, and to represent every point by a position vector from the origin. Vector equality is what makes the whole "slide it wherever you like" freedom of vector algebra legitimate.
Vector equality is a foundational definition in the NCERT Class 12 Vector Algebra chapter, frequently tested through short conceptual CBSE board questions that distinguish it from coinitial or collinear vectors. Students revising "vector algebra class 12 important questions" for boards or JEE Main should treat this component-matching test as a quick sanity check before any vector proof.
Idea: Two matrices are equal iff corresponding entries are equal, so read off one equation per position.
(1,1):x+3=0⇒x=−3
(1,2):z+4=6⇒z=2
(1,3):2y−7=3y−2⇒−5=y⇒y=−5
(2,2):a−1=−3⇒a=−2
(2,3):0=2c+2⇒c=−1
(3,1):b−3=2b+4⇒−7=b⇒b=−7
The remaining positions (−6=−6, −21=−21, 0=0) are automatically satisfied.
✓Final answer
a=−2,b=−7,c=−1,x=−3,y=−5,z=2.
Equating the two matrices entry by entry gives a=−2,b=−7,c=−1,x=−3,y=−5,z=2.
Two matrices of the same order are equal exactly when every entry in the same position matches. So this single matrix equation splits into nine ordinary equations; six carry the unknowns and three are automatically true.
(1,3):2y−7=3y−2. Bring terms together: −7+2=3y−2y, so −5=y, i.e. y=−5.
(2,1):−6=−6 — always true.
(2,2):a−1=−3⇒a=−2.
(2,3):0=2c+2⇒2c=−2⇒c=−1.
(3,1):b−3=2b+4. Then −3−4=2b−b, so b=−7.
(3,2):−21=−21 — always true.
(3,3):0=0 — always true.
All nine equations are consistent, so every unknown is determined.
Watch out
Mind the signs when rearranging: 2y−7=3y−2 gives y=−5 (not +5), and b−3=2b+4 gives b=−7.
✓Final answer
a=−2,b=−7,c=−1,x=−3,y=−5,z=2.
Method: Solving unknowns from equality of two matrices
Use this whenever two matrices are set equal and you must find the unknowns inside them.
Steps
Step 1: Use the equality condition.
Two matrices of the same order are equal iff every corresponding entry is equal. This turns one matrix equation into a set of scalar equations, one per position.
Step 2: Write down each entry equation.
Match position by position. Some positions give trivially true statements (e.g. −6=−6) and can be skipped; the rest are equations in the unknowns.
Step 3: Solve each equation, watching the signs.
Many are one-line linear equations; isolate each unknown and solve.
Common Mistakes
Mistake 1: Sign errors when rearranging.
Why it's wrong: 2y−7=3y−2 gives y=−5 (not +5), and b−3=2b+4 gives b=−7. Correct approach: move variables to one side and constants to the other, tracking each sign.
Mistake 2: Matching entries in the wrong positions.
Why it's wrong: equality is position-by-position; comparing (1,3) with (3,1) produces false equations. Correct approach: equate only entries in identical (row, column) positions.
Mistake 3: Assuming an unknown appears where it doesn't.
Why it's wrong: some positions are pure constants and give no information about the unknowns. Correct approach: extract equations only from positions that actually contain an unknown.