The points with position vectors 60î + 3ĵ, 40î -8ĵ, aî - 52ĵ are collinear if a is equal to

Question

The points with position vectors 60&icirc; + 3&jcirc;, 40&icirc; -8&jcirc;, a&icirc; - 52&jcirc; are collinear if a is equal to

Prepp Team · Accepted Answer

Finding the Value for Collinear Points using Position Vectors

We are given three points with the following position vectors:

Point P: \vec{p_1} = 60\hat{i} + 3\hat{j}
Point Q: \vec{p_2} = 40\hat{i} - 8\hat{j}
Point R: \vec{p_3} = a\hat{i} - 52\hat{j}

These position vectors correspond to the coordinates of the points:

P: (60, 3)
Q: (40, -8)
R: (a, -52)

For three points to be collinear, they must lie on the same straight line. There are several ways to check for collinearity. We can use the vector approach or the slope approach.

Method 1: Using Vector Collinearity

Three points P, Q, and R are collinear if the vector \vec{PQ} is parallel to the vector \vec{QR}. This means \vec{PQ} = k\vec{QR} for some scalar k.

First, let's find the vectors \vec{PQ} and \vec{QR}.

The vector \vec{PQ} is the difference between the position vector of Q and the position vector of P:

\vec{PQ} = \vec{p_2} - \vec{p_1} = (40\hat{i} - 8\hat{j}) - (60\hat{i} + 3\hat{j}) = (40 - 60)\hat{i} + (-8 - 3)\hat{j} = -20\hat{i} - 11\hat{j}

The vector \vec{QR} is the difference between the position vector of R and the position vector of Q:

\vec{QR} = \vec{p_3} - \vec{p_2} = (a\hat{i} - 52\hat{j}) - (40\hat{i} - 8\hat{j}) = (a - 40)\hat{i} + (-52 - (-8))\hat{j} = (a - 40)\hat{i} + (-52 + 8)\hat{j} = (a - 40)\hat{i} - 44\hat{j}

For \vec{PQ} and \vec{QR} to be parallel, the ratio of their corresponding components must be equal:

\frac{\text{i-component of }\vec{PQ}}{\text{i-component of }\vec{QR}} = \frac{\text{j-component of }\vec{PQ}}{\text{j-component of }\vec{QR}}

\frac{-20}{a - 40} = \frac{-11}{-44}

Simplify the ratio on the right side:

\frac{-11}{-44} = \frac{11}{44} = \frac{1}{4}

So, the equation becomes:

\frac{-20}{a - 40} = \frac{1}{4}

Now, we can cross-multiply to solve for a:

-20 \times 4 = 1 \times (a - 40)

-80 = a - 40

Add 40 to both sides:

a = -80 + 40

a = -40

Method 2: Using Slopes

Three points (x₁, y₁), (x₂, y₂), and (x₃, y₃) are collinear if the slope of the line segment connecting (x₁, y₁) and (x₂, y₂) is equal to the slope of the line segment connecting (x₂, y₂) and (x₃, y₃).

The formula for the slope between two points (x_a, y_a) and (x_b, y_b) is \frac{y_b - y_a}{x_b - x_a}.

Points are P(60, 3), Q(40, -8), and R(a, -52).

Slope of PQ:

\text{Slope}_{PQ} = \frac{-8 - 3}{40 - 60} = \frac{-11}{-20} = \frac{11}{20}

Slope of QR:

\text{Slope}_{QR} = \frac{-52 - (-8)}{a - 40} = \frac{-52 + 8}{a - 40} = \frac{-44}{a - 40}

For collinearity, Slope_PQ = Slope_QR:

\frac{11}{20} = \frac{-44}{a - 40}

Cross-multiply:

11 \times (a - 40) = 20 \times (-44)

11a - 440 = -880

Add 440 to both sides:

11a = -880 + 440

11a = -440

Divide by 11:

a = \frac{-440}{11}

a = -40

Both methods yield the same value for a.

Conclusion

For the given points with position vectors 60\hat{i} + 3\hat{j}, 40\hat{i} -8\hat{j}, and a\hat{i} - 52\hat{j} to be collinear, the value of a must be -40.

Concept	Description	Condition for Collinearity
Position Vector	A vector representing the position of a point relative to the origin. \vec{r} = x\hat{i} + y\hat{j} for point (x, y).	N/A (Defines points)
Collinear Points	Three or more points that lie on the same straight line.	\vec{PQ} = k\vec{QR} for points P, Q, R (vector method) OR Slope_PQ = Slope_QR (slope method) OR Area of triangle PQR = 0.
Vector between two points	For points P with position vector \vec{p} and Q with position vector \vec{q}, the vector \vec{PQ} = \vec{q} - \vec{p}.	Used to form vectors \vec{PQ} and \vec{QR}.

Revision Table: Collinear Points & Vectors

Term	Definition	Application in Problem
Position Vector	Vector from origin to a point. E.g., (x, y) is represented by x\hat{i} + y\hat{j}.	Used to get coordinates of the three points.
Vector between Points	Vector from point A to point B: \vec{AB} = \vec{b} - \vec{a}.	Calculated \vec{PQ} and \vec{QR}.
Collinearity (Vector)	Vectors \vec{AB} and \vec{BC} are parallel (one is a scalar multiple of the other).	\vec{PQ} and \vec{QR} components are proportional: \frac{-20}{a-40} = \frac{-11}{-44}.
Collinearity (Slope)	Slope between A and B equals slope between B and C.	\frac{y_2-y_1}{x_2-x_1} = \frac{y_3-y_2}{x_3-x_2}. Used \frac{11}{20} = \frac{-44}{a-40}.

Additional Information: Understanding Collinearity

Collinearity is a fundamental concept in geometry and vector algebra. When points are collinear, they lie on a single straight line. This implies a linear relationship between their coordinates.

In the context of vectors, if three points P, Q, and R are collinear, then the vector \vec{PR} can be expressed as a scalar multiple of \vec{PQ} (i.e., \vec{PR} = k\vec{PQ} for some scalar k). This is because the vectors lie along the same direction (or opposite direction if k is negative).

Alternatively, using the method applied in the solution, if P, Q, and R are collinear, then the vectors \vec{PQ} and \vec{QR} are parallel. Since they share a common point Q, being parallel forces them to lie on the same line, making P, Q, and R collinear.

The slope method is essentially a specific case of the vector method in 2D coordinates, where the slope is the ratio of the y-component to the x-component of the vector connecting two points. Setting slopes equal is equivalent to setting the ratio of components equal, which is the condition for parallel vectors.

Another common method is to check if the area of the triangle formed by the three points is zero. If the points are collinear, they cannot form a triangle with any area.