Q.Show that the points $P(-2, 3, 5)$, $Q(1, 2, 3)$ and $R(7, 0, -1)$ are collinear.

The key idea: the XY-plane is simply the set of all points where $z=0$. When a plane divides a line segment, the point of intersection lies on both the line and the plane. So we need the point on line AB whose $z$-coordinate is zero, and then find the ratio in which that point divides AB.

The section formula for three dimensions works exactly like the two-dimensional version, applied separately to each coordinate. If a point P divides AB in the ratio $k:1$ (with sign indicating internal or external), then its coordinates are:

for internal division. For external division, the formula uses $k:1$ with $k$ negative, or equivalently we can use $k:-1$.

1. Set up the condition. Let the point where the line meets the XY-plane be P. Since P lies on AB, we can say P divides AB in some ratio $\lambda : 1$ (where $\lambda$ can be positive for internal, negative for external). Then the $z$-coordinate of P is:

2. Apply the plane condition. On the XY-plane, $z=0$. So:

This gives $-3\lambda - 5 = 0$, so $\lambda = -\frac{5}{3}$.

3. Interpret the ratio. The ratio is $\lambda : 1 = -\frac{5}{3} : 1$. Multiply through by 3 to get $-5 : 3$. A negative sign in the ratio means the division is external — the point lies outside the segment AB, closer to B. The magnitude $5:3$ tells us the distances: AP : PB = 5 : 3 externally.

4. Verify quickly. If the ratio is $5:3$ externally, then the point divides AB such that AP : PB = 5 : 3, with P on the extension beyond B (since the ratio is negative). Check: from A to B, $z$ goes from $-5$ to $-3$ (increasing). The external point with $z=0$ lies beyond B, which makes sense because $z$ would continue increasing past $-3$ to $0$.

The key idea: the image of a point in a line is the point such that the line is the perpendicular bisector of the segment joining the point and its image. So the foot of the perpendicular from the point to the line is the midpoint of the point and its image. Find that foot, then double it.

1. Write the line in parametric form. The line is $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda$ (say). So any point on the line is:

2. Let the foot of the perpendicular from $A(1,6,3)$ to the line be $F$.

   $F$ lies on the line, so $F = (\lambda,\; 2\lambda+1,\; 3\lambda+2)$ for some $\lambda$.

3. The vector $\overrightarrow{AF}$ must be perpendicular to the direction vector of the line.

   Direction vector of the line: $\vec{d} = (1,2,3)$.

   Vector $\overrightarrow{AF} = (\lambda-1,\; (2\lambda+1)-6,\; (3\lambda+2)-3) = (\lambda-1,\; 2\lambda-5,\; 3\lambda-1)$

   Perpendicular condition: $\overrightarrow{AF} \cdot \vec{d} = 0$

4. So the foot $F$ is at $\lambda = 1$:

5. Now use the midpoint property. If $A'(x,y,z)$ is the image of $A$ in the line, then $F$ is the midpoint of $AA'$:

Equating coordinates:

• $\frac{1+x}{2} = 1 \Rightarrow 1+x = 2 \Rightarrow x = 1$
• $\frac{6+y}{2} = 3 \Rightarrow 6+y = 6 \Rightarrow y = 0$
• $\frac{3+z}{2} = 5 \Rightarrow 3+z = 10 \Rightarrow z = 7$

So the image is $(1,0,7)$.

Step 1 — Read off the line and the plane.

The line $\dfrac{x-4}{1}=\dfrac{y-2}{1}=\dfrac{z-k}{2}$ passes through the point $A(4,2,k)$ with direction $\vec b=(1,1,2)$.

The plane $2x-4y+z=7$ has normal $\vec n=(2,-4,1)$.

Step 2 — Condition (i): the line must be parallel to the plane.

If the line lies in the plane, it certainly cannot pierce it, so its direction must be perpendicular to the normal:

This is satisfied for every $k$ (the direction ratios do not involve $k$), so it only tells us the line is parallel to the plane. It is not yet in it — a parallel line could sit above it.

Step 3 — Condition (ii): a point of the line must lie in the plane.

Substitute $A(4,2,k)$ into $2x-4y+z=7$:

Step 4 — Verify.

With $k=7$ the line is $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}=t$, i.e. $(4+t,\,2+t,\,7+2t)$. Substituting into the plane:

True for all $t$, confirming that the whole line lies in the plane.

ANSWER: D