Analysing the Points on the Line Joining A and B

We are given the position vectors of two points, A and B, and asked to determine which of the given points lie on the line joining A and B.

The position vector of point A is $\vec{a} = \hat{i} - \hat{j}$. The coordinates of A are $(1, -1, 0)$.

The position vector of point B is $\vec{b} = \hat{j} + \hat{k}$. The coordinates of B are $(0, 1, 1)$.

Finding the Direction Vector of the Line AB

The vector representing the direction of the line joining A and B can be found by calculating the vector $\vec{AB}$:

$\vec{AB} = \vec{b} - \vec{a}$

$\vec{AB} = (\hat{j} + \hat{k}) - (\hat{i} - \hat{j})$

$\vec{AB} = 0\hat{i} + 1\hat{j} + 1\hat{k} - 1\hat{i} + 1\hat{j} - 0\hat{k}$

$\vec{AB} = (0-1)\hat{i} + (1+1)\hat{j} + (1-0)\hat{k}$

$\vec{AB} = -\hat{i} + 2\hat{j} + \hat{k}$

Equation of the Line Joining A and B

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{v}$ is given by $\vec{r} = \vec{a} + t\vec{v}$, where $t$ is a scalar parameter.

Here, the line passes through point A (with position vector $\vec{a}$) and is parallel to the vector $\vec{AB}$.

So, the equation of the line joining A and B is:

$\vec{r} = \vec{a} + t\vec{AB}$

$\vec{r} = (\hat{i} - \hat{j}) + t(-\hat{i} + 2\hat{j} + \hat{k})$

$\vec{r} = \hat{i} - \hat{j} - t\hat{i} + 2t\hat{j} + t\hat{k}$

$\vec{r} = (1-t)\hat{i} + (-1+2t)\hat{j} + t\hat{k}$

Any point (x, y, z) on this line will have a position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. By comparing the coefficients, we get the parametric equations of the line:

• $x = 1-t$
• $y = -1+2t$
• $z = t$

Checking the Given Points

We will now check if each of the given points satisfies the parametric equations for some value of $t$.

Checking Point 1: (-1, -3, 1)

Let's assume this point lies on the line. Then its coordinates must satisfy the parametric equations:

• $-1 = 1-t$
• $-3 = -1+2t$
• $1 = t$

From the third equation, $t = 1$.

Substitute $t=1$ into the first equation: $-1 = 1 - 1 = 0$. This is false ($-1 \neq 0$).

Since the equations are not consistent for a single value of $t$, point 1 does not lie on the line joining A and B.

Checking Point 2: (-1, 3, 2)

Let's assume this point lies on the line. Then its coordinates must satisfy the parametric equations:

• $-1 = 1-t$
• $3 = -1+2t$
• $2 = t$

From the third equation, $t = 2$.

Substitute $t=2$ into the first equation: $-1 = 1 - 2 = -1$. This is true.

Substitute $t=2$ into the second equation: $3 = -1 + 2(2) = -1 + 4 = 3$. This is true.

Since all three equations are satisfied for $t = 2$, point 2 lies on the line joining A and B.

Checking Point 3: (-2, 5, 3)

Let's assume this point lies on the line. Then its coordinates must satisfy the parametric equations:

• $-2 = 1-t$
• $5 = -1+2t$
• $3 = t$

From the third equation, $t = 3$.

Substitute $t=3$ into the first equation: $-2 = 1 - 3 = -2$. This is true.

Substitute $t=3$ into the second equation: $5 = -1 + 2(3) = -1 + 6 = 5$. This is true.

Since all three equations are satisfied for $t = 3$, point 3 lies on the line joining A and B.

Conclusion on Points Lying on the Line

Based on our checks, point 1 does not lie on the line joining A and B, while points 2 and 3 do lie on the line.

Revision Table: Point Check Summary

Therefore, points 2 and 3 lie on the line joining A and B.

Additional Information: Vector Equation Forms

The equation of a line in 3D space can be expressed in different forms:

• Vector Form: $\vec{r} = \vec{a} + t\vec{v}$, where $\vec{a}$ is the position vector of a known point on the line and $\vec{v}$ is a vector parallel to the line.
• Parametric Form: If $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, $\vec{a} = x_0\hat{i} + y_0\hat{j} + z_0\hat{k}$, and $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$, the parametric equations are $x = x_0 + ta$, $y = y_0 + tb$, $z = z_0 + tc$. These are what we used to check the points.
• Cartesian Form: By eliminating the parameter $t$ from the parametric equations, we get the Cartesian form: $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ (provided $a, b, c \neq 0$). If a direction cosine is zero, say $a=0$, then the equations become $x = x_0$ and $\frac{y - y_0}{b} = \frac{z - z_0}{c}$. For our line AB, $x_0=1, y_0=-1, z_0=0$ and direction ratios are $a=-1, b=2, c=1$. The Cartesian form is $\frac{x - 1}{-1} = \frac{y - (-1)}{2} = \frac{z - 0}{1}$, which simplifies to $\frac{x - 1}{-1} = \frac{y + 1}{2} = \frac{z}{1}$. A point (x, y, z) lies on the line if it satisfies these ratios.