## Position of a Vector

If we are provided with a point Q(x,y,z) and \(\overline{OQ}(=\underset{r}{\rightarrow})=x\widehat{i}+y\widehat{j}+z\widehat{k}\) and the magnitude is given by \(\sqrt{x^{2}+y^{2}+z^{2}}\). The direction ratios for a vector is its scalar components and is responsible for its projections along the respective axes.

### The relation between magnitude, direction ratios, direction cosines of a vector

If a vector has been given with dimensions such as magnitude(p), direction ratios (x,y,z) and direction cosines (l,m,n) then the relation between them is:

l=x/p, m=y/p,n=z/p

The order taken for the vector sum of the three sides of the triangle is \(\underset{0}{\rightarrow}\)

And the vector sum of coinitial vectors is the diagonal of the parallelogram which has the vectors as its adjacent sides.

When multiplying a vector by a scalar λ, the magnitude of the vector changes by the multiple |λ|, and the direction remains same (or makes it opposite) according to as the value of λ is positive (or negative).

### Position Vector of a Point **R** on a Line Segment

A line segment joining two points such as P having a position vector of \(\underset{a}{\rightarrow}\) and Q having a position vector of \(\underset{b}{\rightarrow}\) is divided by a point R by the ratio m;n then,

- The internal ratio is given by \(n\underset{a}{\rightarrow}+m\underset{b}{\rightarrow}/m+n\)
- The external ratio is given by

The scalar product of two position vectors \(\underset{a}{\rightarrow}\) and \(\underset{b}{\rightarrow}\) and subtending an angle \(\theta\) is represented by \(\underset{a}{\rightarrow}\).\(\underset{b}{\rightarrow}\)=\(\left | \underset{a}{\rightarrow} \right |\left | \underset{b}{\rightarrow} \right |cos\theta\)

We can find the value of \(\theta\) from the above equation.

Also Access |

NCERT Solutions for Class 12 Maths Chapter 10 |

NCERT Exemplar for Class 12 Maths Chapter 10 |

### Important Questions

- Find the ratio in which B divides AC and also show that the points A(2, – 3, – 9), B(6, 0, –1) and C(10, 4, 6) are not collinear
- Determine the girl’s displacement from her initial point of departure if she walks 5 km towards west, then she walks 4 km in a direction 40° east of north and stops.
- Draw a unit vector in the XY-plane, making an angle of 30° with the positive direction of x-axis
- Find the area of the triangle with vertices P(2, 3, 4), Q(3, 4, 6) and R(2, 6, 7).
- Show that the points P(1, 2, 7), Q(2, 6, 3) and R(3, 10, –1) are collinear

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