(Assertion-Reason) Assertion (A) : A line through the points (4, 7, 8) and (2, 3, 4) is parallel to a line thr

Question

(4, 7, 8)

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

(d) Assertion (A) is false and Reason (R) is true.

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★

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Concept understanding — Cross Product Parallel Vectors

Cross Product of Parallel Vectors

Imagine you're trying to open a door. You push on the handle — that force works because it's perpendicular to the door. If you push along the door (parallel to its surface), nothing happens. The cross product measures exactly this "perpendicular effectiveness" between two vectors.

When two vectors are parallel, they point in exactly the same direction (or exactly opposite). There is no "perpendicular component" between them, so the cross product — which captures that perpendicular interaction — must be zero.

The Intuition

Take two parallel vectors $\vec{a}$ and $\vec{b}$ , two arrows lying along the same line. No matter how you rotate them, you cannot get one to point "across" the other. The area of the parallelogram they span is zero — a degenerate, flat shape. The cross product gives the vector perpendicular to both, with magnitude equal to that area. Since the area is zero, the cross product is the zero vector.

Note

This is why the cross product is called the vector product — its magnitude is $|\vec{a}||\vec{b}|\sin\theta$ , and $\sin\theta = 0$ when $\theta = 0^\circ$ or $180^\circ$ .

The Precise Statement

If $\vec{a}$ and $\vec{b}$ are parallel (i.e. $\vec{b} = k\vec{a}$ for some scalar $k$ ), then:

\vec{a} \times \vec{b} = \vec{0}

The converse is also true: if the cross product of two non-zero vectors is zero, they must be parallel (or anti-parallel).

\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b} \quad (\text{for non-zero vectors})

Why This Matters in Exams

This is a quick check for parallelism: compute a cross product and get zero, and you immediately know the vectors are collinear. It's also used in proofs — for example, showing two lines are parallel by taking the cross product of their direction vectors.

Watch out

A common mistake is to think $\vec{a} \times \vec{b} = \vec{0}$ means $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$ . That's false — it only means they are parallel (or one is zero). The zero vector is parallel to every vector, but the interesting case is when both are non-zero.

Quick Example …

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Two lines are parallel when their direction vectors are scalar multiples of each other. The direction vectors of both given line pairs are $\langle -2, -4, -4 \rangle$ and $\langle 2, 4, 4 \rangle$ , which are scalar multiples, so Assertion (A) is true. Reason (R) is false because the condition for parallel lines is $\vec{b_1} \times \vec{b_2} = \vec{0}$ , not $\vec{b_1} \cdot \vec{b_2} = 0$ . The correct option is (c).

Concept First: What Makes Two Lines Parallel?

In 3D geometry, two lines are parallel if their direction vectors are scalar multiples of each other. That is, if one direction vector can be written as $k$ times the other, for some non-zero scalar $k$ . This is equivalent to saying their cross product is the zero vector: $\vec{b_1} \times \vec{b_2} = \vec{0}$ .

The dot product condition $\vec{b_1} \cdot \vec{b_2} = 0$ means the vectors are perpendicular, not parallel. That's a classic trap — Reason (R) states exactly this wrong condition.

Let's check the Assertion first.

Step-by-Step Solution

1. Find the direction vector of the first line.

The line passes through $(4, 7, 8)$ and $(2, 3, 4)$ . The direction vector is the difference between these points:

\vec{b_1} = (2 - 4, \; 3 - 7, \; 4 - 8) = (-2, -4, -4)

2. Find the direction vector of the second line.

The line passes through $(-1, -2, 1)$ and $(1, 2, 5)$ . Its direction vector is:

\vec{b_2} = (1 - (-1), \; 2 - (-2), \; 5 - 1) = (2, 4, 4)

3. Check if they are parallel.

Observe that $\vec{b_2} = -1 \cdot \vec{b_1}$ :

(2, 4, 4) = -1 \cdot (-2, -4, -4)

Since one is a scalar multiple of the other, the two lines are parallel. Assertion (A) is true.

Tip

You don't always need to compute the scalar factor explicitly. Just check if the ratios of corresponding components are equal: $\frac{2}{-2} = \frac{4}{-4} = \frac{4}{-4} = -1$ . If all three ratios match, the vectors are parallel. …

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

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. If |vector a| = 5 units then |vector a \times vector a| is equal to:$
(a) 5 units
(b) 25 units
(c) 1 units
(d) 0 units
›Reveal solutionSolution

The cross product of any vector with itself is always the zero vector, because the angle between a vector and itself is $0$ .

For any vector $\vec a$ ,

$\vec a \times \vec a = |\vec a||\vec a|\sin(0^\circ)\,\hat n = 0$

…

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CBSE 2025Set E1 markMCQ
$2, 7, 9$
(a) $9$
(b) $18$
(c) $27$
(d) $3$
›Reveal solutionSolution

Proportional DRs: $\frac{2}{6}=\frac{7}{21}=\frac{9}{x}=\frac13\Rightarrow x=27$ .

Parallel lines have direction ratios in proportion:

$\frac{2}{6}=\frac{7}{21}=\frac{9}{x}.$

…

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CBSE 2025Set E1 markMCQ
$a, b, c$
(a) $cy$
(b) $cx$
(c) $bz$
(d) $ax$
›Reveal solutionSolution

Proportional DRs: $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$ , so cross-multiplying the first and third gives $az=cx$ .

For parallel lines the direction ratios are proportional:

$\frac{a}{x}=\frac{b}{y}=\frac{c}{z}.$

…

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CBSE 2025Set ANNUAL1 markMCQ
$\lambda$
(a) $\dfrac{2}{3}$
(b) $\dfrac{3}{2}$
(c) $\dfrac{5}{2}$
(d) $\dfrac{2}{5}$
›Reveal solutionSolution

Parallel vectors have proportional components; match the third component's ratio to the other two.

For $3\hat i - 6\hat j + \hat k$ and $2\hat i - 4\hat j + \lambda\hat k$ to be parallel, corresponding components must be in the same ratio:

$\dfrac{3}{2} = \dfrac{-6}{-4} = \dfrac{1}{\lambda}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If a⃗ is a non-zero vector then |a⃗ \times a⃗| is equal to:$
(a) |a⃗|
(b) |a⃗|²
(c) 1
(d) 0
›Reveal solutionSolution

The cross product of any vector with itself is always the zero vector, so its magnitude is $0$ .

For any non-zero vector $\vec{a}$ , the angle between $\vec{a}$ and itself is $\theta = 0$ . Using $\vec{a}\times\vec{a} = |\vec{a}||\vec{a}|\sin\theta\,\hat n$ , and $\sin 0 = 0$ , the whole expres …

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CBSE 2024Set A11 mark
$K$
›Reveal solutionSolution

Proportional direction ratios for parallel lines give $\dfrac{3}{K}=\dfrac{1}{2}$ , hence $K=6$ .

Two lines are parallel iff their direction ratios are proportional. With ratios $1,3,5$ and $2,K,10$ ,

$\frac{1}{2}=\frac{3}{K}=\frac{5}{10}.$ …
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CBSE 2024Set D1 markMCQ
$\frac{x-3}{a} = \frac{y-4}{b} = \frac{z-5}{c}$
(a) $5a + 3b + 2c = 0$
(b) $\frac{a}{5} = \frac{b}{3} = \frac{c}{2}$
(c) $5a = 3b = 2c$
(d) none of these
›Reveal solutionSolution

Parallel lines $\Rightarrow$ direction ratios proportional: $\frac{a}{5} = \frac{b}{3} = \frac{c}{2}$ .

The first line has direction ratios $(a, b, c)$ and the second has $(5, 3, 2)$ . Two lines are parallel iff their direction ratios are proportional:

$\frac{a}{5} = \frac{b}{3} = \frac{c}{2}.$ …

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CBSE 2023Set 65/3/11 markMCQ
$(4, 7, 8)$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
›Reveal solutionSolution

Two lines are parallel when their direction vectors are scalar multiples of each other. The direction vectors of both given line pairs are $\langle -2, -4, -4 \rangle$ and $\langle 2, 4, 4 \rangle$ , which are scalar multiples, so Assertion (A) is true. Reason (R) is false because the condition for parallel lines is $\vec{b_1} \times \vec{b_2} = \vec{0}$ , not $\vec{b_1} \cdot \vec{b_2} = 0$ . The correct option is (c).

Concept First: What Makes Two Lines Parallel?

In 3D geometry, two lines are parallel if their direction vectors are scalar multiples of each other. That is, if one direction vector can be written as $k$ times the other, for some non-zero scalar $k$ . This is equivalent to saying their cross product is the zero vector: $\vec{b_1} \times \vec{b_2} = \vec{0}$ .

The dot product condition $\vec{b_1} \cdot \vec{b_2} = 0$ means the vectors are perpendicular, not parallel. That's a classic trap — Reason (R) states exactly this wrong condition.

Let's check the Assertion first.

Step-by-Step Solution

1. Find the direction vector of the first line.

The line passes through $(4, 7, 8)$ and $(2, 3, 4)$ . The direction vector is the difference between these points:
$\vec{b_1} = (2 - 4, \; 3 - 7, \; 4 - 8) = (-2, -4, -4)$
2. Find the direction vector of the second line.

The line passes through $(-1, -2, 1)$ and $(1, 2, 5)$ . Its direction vector is:
$\vec{b_2} = (1 - (-1), \; 2 - (-2), \; 5 - 1) = (2, 4, 4)$
3. Check if they are parallel.

Observe that $\vec{b_2} = -1 \cdot \vec{b_1}$ :
$(2, 4, 4) = -1 \cdot (-2, -4, -4)$
Since one is a scalar multiple of the other, the two lines are parallel. Assertion (A) is true.

Tip

You don't always need to compute the scalar factor explicitly. Just check if the ratios of corresponding components are equal: $\frac{2}{-2} = \frac{4}{-4} = \frac{4}{-4} = -1$ . If all three ratios match, the vectors are parallel. …

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CBSE 2023Set E1 markMCQ
$x,5,3$
(a) $10$
(b) $5$
(c) $3$
(d) $40$
›Reveal solutionSolution

Proportional direction ratios give $x=10$ .

Parallel lines have proportional direction ratios, so

$\frac{x}{20}=\frac{5}{10}=\frac{3}{6}.$

…

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CBSE 2023Set E1 markMCQ
$a_{1},b_{1},c_{1}$
(a) $b_{1}$
(b) $b_{2}$
(c) $b_{3}$
(d) $c_{1}$
›Reveal solutionSolution

From proportional direction ratios, $\dfrac{a_1c_2}{a_2}=c_1$ .

For parallel lines the direction ratios are proportional:

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.$

…

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CBSE 2022Set ANNUAL1 markMCQ
$3\vec a \times 4\vec a = ?$
(a) $0$
(b) $12$
(c) $12\vec a$
(d) $\vec 0$
›Reveal solutionSolution

The cross product of any vector with itself (or a scalar multiple of itself) is the zero vector.

$3\vec a\times4\vec a=12(\vec a\times\vec a)$ .

…

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CBSE 2020Set ANNUAL1 mark
$p$
›Reveal solutionSolution

Two vectors are parallel exactly when their corresponding components are proportional; matching ratios gives $p$ .

Given vectors $3\hat{i}+2\hat{j}+9\hat{k}$ and $\hat{i}+p\hat{j}+3\hat{k}$ are parallel, so their components must be proportional:

$\frac{3}{1} = \frac{2}{p} = \frac{9}{3}$

…

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Concept First: What Makes Two Lines Parallel?

Step-by-Step Solution

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