Q.Show that the line through the points $(4, 7, 8), (2, 3, 4)$ is parallel to the line through the points $(-1, -2, 1), (1, 2, 5)$.

Concept: Two lines are parallel if their direction vectors are scalar multiples of each other.

Step 1 – Direction vector of first line

From $(4,7,8)$ to $(2,3,4)$:

$\vec{d_1} = (2-4,\ 3-7,\ 4-8) = (-2, -4, -4)$

Step 2 – Direction vector of second line

From $(-1,-2,1)$ to $(1,2,5)$:

$\vec{d_2} = (1-(-1),\ 2-(-2),\ 5-1) = (2, 4, 4)$

Step 3 – Check scalar multiple

$\vec{d_1} = (-2, -4, -4) = -1 \cdot (2, 4, 4) = -1 \cdot \vec{d_2}$

Since $\vec{d_1} = k \vec{d_2}$ with $k = -1$, the direction vectors are parallel.

Why direction vectors decide parallelism

In 3D geometry, a line is completely determined by a point on it and a direction vector — the vector that points from one point on the line to another. Two lines are parallel precisely when their direction vectors are scalar multiples of each other. That is, if one direction vector can be multiplied by some constant (positive, negative, or even a fraction) to get the other, the lines run in the same or exactly opposite directions — and that’s the definition of parallel lines in space.

The actual positions of the points don’t matter for parallelism; only the direction matters. So we ignore the given points themselves and focus on the vectors connecting each pair.

Step-by-step solution

1. Find the direction vector of the first line.

The first line passes through $A(4, 7, 8)$ and $B(2, 3, 4)$. The direction vector $\vec{d_1}$ is simply $B - A$:

2. Find the direction vector of the second line.

The second line passes through $C(-1, -2, 1)$ and $D(1, 2, 5)$. Its direction vector $\vec{d_2}$ is $D - C$:

3. Check if the two vectors are scalar multiples.

We ask: does there exist a scalar $k$ such that $\vec{d_1} = k \cdot \vec{d_2}$? Compare component by component:

From the first component: $-2 = k \cdot 2 \implies k = -1$.

Check the second: $-4 = (-1) \cdot 4 = -4$ — works.

Check the third: $-4 = (-1) \cdot 4 = -4$ — works.

So $\vec{d_1} = -1 \cdot \vec{d_2}$. The scalar $k = -1$ is a real number, so the condition is satisfied.

4. Conclude about the lines.

Since the direction vectors are scalar multiples, the two lines are parallel. The fact that $k$ is negative simply means they point in opposite directions — but in geometry, opposite directions are still parallel.

Method: Parallelism of lines via scalar-multiple direction vectors

Two lines are parallel exactly when their direction vectors point the same way or exactly opposite — that is, one is a scalar multiple of the other. The multiplier may be any non-zero real, positive or negative.

Steps

Step 1: Build each direction vector from the two given points of each line:

Step 2: Test for a common scalar. Solve $\vec d_1=k\,\vec d_2$ component-wise. Read $k$ from the first component, then confirm the same $k$ works for all three:

Equivalently, $\vec d_1\times\vec d_2=\vec0$.

Step 3: Interpret the sign of $k$. A negative $k$ (opposite sense) is still parallel — parallelism cares about the line, not its arrow. Reversing which endpoint you subtract first just flips $k$'s sign.

Step 4: Conclude parallel if a single consistent $k$ exists; if the ratios disagree, the lines are not parallel.

Common Mistakes

Mistake 1: Declaring the lines "not parallel" because $\vec d_1\ne\vec d_2$.

Why it's wrong: parallelism only needs one vector to be a scalar multiple of the other, not equal. Correct approach: here $\vec d_1=(-2,-4,-4)=-1\cdot(2,4,4)=-\vec d_2$, so they are parallel.

Mistake 2: Rejecting a negative scalar.

Why it's wrong: $k=-1$ means opposite directions, which is still parallel. Correct approach: accept any single non-zero $k$ that works for all three components.

Mistake 3: Checking only one component's ratio.

Why it's wrong: one matching ratio can be coincidence; parallelism needs the same $k$ across all three. Correct approach: confirm $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.