A vector $\vec A$ is multiplied by (i) a positive scalar $\lambda 1$, (ii) a positive scalar $0

Question

A vector $\vec A$ is multiplied by (i) a positive scalar $\lambda > 1$, (ii) a positive scalar $0 < \lambda < 1$, and (iii) a negative scalar $\lambda < 0$. Explain, in each case, what happens to the magnitude and direction of the resulting vector $\lambda\vec A$.

Accepted Answer

Multiplying $\vec A$ by a real number $\lambda$ gives $\lambda\vec A$ , with magnitude $|\lambda\vec A| = |\lambda|\,A$ . (i) For $\lambda > 1$ (a positive number greater than 1), the direction of $\lambda\vec A$ is unchanged from $\vec A$ , but its magnitude becomes larger than $A$ — the vector is stretched. (ii) For $0 < \lambda < 1$ , the direction again stays the same as $\vec A$ , but the magnitude becomes smaller than $A$ — the vector is shrunk while keeping the same direction. (iii) For $\lambda < 0$ , the magnitude is scaled by $|\lambda|$ exactly as before, but the direction is reversed through $180^\circ$ — the resulting vector points exactly opposite to $\vec A$ . This last case, with $\lambda = -1$ , is what defines the negative of a vector, $-\vec A$ , used to construct vector subtraction. [!ANSWER] (i) same direction, magnitude increases; (ii) same direction, magnitude decreases; (iii) direction reversed, magnitude scaled by $|\lambda|$ .

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2023Set ANNUAL1 mark
$Q. When a vector A is multiplied by a negative number λ, what will you get?$
›Reveal solutionSolution
lambda A (lambda negative) has magnitude |lambda||A| and points opposite to A.

When a vector A is multiplied by a real number lambda, the result lambda A is itself a vector.

Its magnitude becomes |lambda| times the magnitude of A: |lambda A| = |lambda| x |A|.

Its direction depends on the sign of lambda:

If lambda is positive, lambda A points in the SAME direction as A (just scaled in length).

If lambda is negative, lambda A points in the OPPOSITE direction to A (reversed by 180 degrees), and its magnitude is scaled by |lambda|.

So for a negative lambda, the resulting vector lambda A is a vector antiparallel to A, with magnitude |lambda| times that of A. For example, -2A is a vector twice as long as A but pointing exactly opposite to it.

✓Final answer
$lambda A (with lambda negative) is a vector of magnitude |lambda| times |A|, pointing in the direction exactly opposite to A.$
CBSE 2022Set ANNUAL1 mark
$Q. Show that vectors A = 2i^ - 3j^ + 4k^ and B = -6i^ + 9j^ - 12k^ are parallel to each other.$
›Reveal solutionSolution
B works out to exactly −3 times A component-by-component, confirming the two vectors lie along the same line — they are parallel (specifically anti-parallel, pointing in opposite senses).

Step 1: Write down the given vectors.

A = 2i^ − 3j^ + 4k^

B = −6i^ + 9j^ − 12k^

Step 2: Test whether B is a scalar multiple of A.

Compare each component's ratio:

i^ component: −6/2 = −3

j^ component: 9/(−3) = −3

k^ component: −12/4 = −3

Step 3: Confirm the ratio is the same, λ = −3, for every component.

B = −3(2i^ − 3j^ + 4k^) = −3A

Step 4: Interpret the result.

Two vectors are parallel to each other whenever one equals a scalar multiple of the other, i.e. B = λA for some real λ. Here λ = −3 is negative, meaning B points in exactly the opposite direction to A but lies along the same straight line — so A and B are indeed parallel vectors (anti-parallel).

Step 5: Check magnitude consistency (optional verification).

|A| = √(2^2+3^2+4^2) = √29

|B| = √(6^2+9^2+12^2) = √(36+81+144) = √261 = √(9×29) = 3√29 = 3|A| ✓ — consistent with B = −3A.

✓Final answer
$Since B = -3A, vector B is a scalar multiple of A, proving A and B are parallel (anti-parallel) vectors.$

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Q.A vector $\vec A$ is multiplied by

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