Vectors: Position Vectors and Direction Cosines | NIMCET

Quick answer: A vector has magnitude and direction. For $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, the magnitude is $\lvert\vec{a}\rvert = \sqrt{a_1^2 + a_2^2 + a_3^2}$ and the unit vector is $\hat{a} = \frac{\vec{a}}{\lvert\vec{a}\rvert}$. The position vector of a point $P(x, y, z)$ is $x\hat{i} + y\hat{j} + z\hat{k}$, and $\overrightarrow{AB} = \vec{b} - \vec{a}$. The direction cosines $l, m, n$ are the cosines of the angles with the axes, and $l^2 + m^2 + n^2 = 1$.

Syllabus note: Several admission and coaching portals report that Vectors was removed in the revised NIMCET mathematics syllabus. Check the official information brochure for your exam year before prioritising this topic. Our revised syllabus update has the details. This article stays useful for older papers and for other exams that include vectors.

Types of vectors

Addition and scalar multiplication

Triangle inequality: $\lvert\vec{a} + \vec{b}\rvert \le \lvert\vec{a}\rvert + \lvert\vec{b}\rvert$.

Position vectors and section formula

With position vectors $\vec{a}$ and $\vec{b}$ for A and B:

This matches the coordinate section formula.

Direction cosines and direction ratios

If a vector makes angles α, β, γ with the x-, y- and z-axes:

$l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma$

and $l^2 + m^2 + n^2 = 1$.

Any numbers $a, b, c$ proportional to $l, m, n$ are direction ratios. To convert:

$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$

For the line through $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the direction ratios are $x_2 - x_1$, $y_2 - y_1$, $z_2 - z_1$.

Since $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$, it follows that $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$.

Worked examples

Example 1: magnitude and unit vector

For $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$, $\lvert\vec{a}\rvert = \sqrt{4 + 1 + 4} = 3$, so

$\hat{a} = \frac{2}{3}\hat{i} - \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}$

These are also the direction cosines.

Example 2: section formula

A and B have position vectors $\hat{i} + 2\hat{j}$ and $4\hat{i} + 8\hat{j}$. Find the point dividing AB internally in $1 : 2$.

$\frac{1(4\hat{i} + 8\hat{j}) + 2(\hat{i} + 2\hat{j})}{3} = 2\hat{i} + 4\hat{j}$

Example 3: collinear vectors

For what λ are $2\hat{i} + 3\hat{j}$ and $\lambda\hat{i} + 6\hat{j}$ collinear?

The components must be proportional: $\frac{\lambda}{2} = \frac{6}{3}$, so $\lambda = 4$.

Example 4: equal angles with the axes

A vector makes equal angles with all three axes. Find its direction cosines.

$3l^2 = 1$, so $l = m = n = \pm\frac{1}{\sqrt{3}}$.

Summary table

Exam tips

• The components of a unit vector are its direction cosines.
• Collinearity means proportional components.
• Use $\overrightarrow{AB} = \vec{b} - \vec{a}$ (head minus tail) every time.

Next, read the dot product and cross product of vectors, or see the NIMCET Coordinate Geometry and Vectors chapter.

Common mistakes

• Writing $\overrightarrow{AB} = \vec{a} - \vec{b}$.
• Treating direction ratios as direction cosines without dividing by the magnitude.
• Swapping m and n in the vector section formula.
• Forgetting that the zero vector has no fixed direction.

Frequently Asked Questions

What is a position vector?

The position vector of a point P is the vector from the origin to P. For P(x, y, z), it is xî + yĵ + zk̂.

What are direction cosines?

They are the cosines of the angles a vector makes with the x-, y- and z-axes. They satisfy l² + m² + n² = 1.

How do you find a unit vector?

Divide the vector by its magnitude.

What is the difference between direction cosines and direction ratios?

Direction ratios are any numbers proportional to the direction cosines. Direction cosines are the normalised values whose squares add to 1.

Are vectors in the NIMCET syllabus?

Several portals report that Vectors was removed in the revised NIMCET syllabus. Confirm with the official information brochure for your exam year.