Vector Solutions, Shortcuts, Formulas

Section Formula

Let P and Q be two points represented by the position vectors OP and OQ , respectively, with respect to the origin O. Then the line segment joining the points P and Q may be divided by a third point, say R, in two ways – internally (Fig 10.16) and externally (Fig 10.17). Here, we intend to find the position vector OR for the point R with respect to the origin O .we take the two cases ;
Case I) where R divided PQ internally . where m and n point R divides PQ internally in the ratio of m : n . Now from triangles ORQ and OPR, we have,
\vec{r}=\frac{m\vec{b}+n\vec{a}}{m+n}
Hence , the position vector of the point R which divides P and Q internally in the ratio of m: n is given b \vec{OR}=\frac{m\vec{b}+n\vec{a}}{m+n}
Case II) When R divides PQ externally (Fig 10.17). We leave it to the reader as an exercise to verify
that the position vector of the point R which divides the line segment PQ externally in the ratio m : n , i.e
\frac{PR}{QR}=\frac{m}{n},\vec{or}=\frac{m\vec{b}-n\vec{a}}{m+n}
Product of two vector : other algebraic operation which we intend to discuss regarding vectors is their product. We may recall that product of two numbers is a number, product of two matrices is again a matrix. But in case of functions, we may multiply them in two ways, namely, multiplication of two functions point wise and composition of two functions. Similarly, multiplication of two vectors is also defined in two ways, namely, scalar (or dot) product where the result is a scalar, and vector (or cross) product where the result is a vector. Based upon these two types of products for vectors, they have found various applications in geometry, mechanics and engineering. In this section, we will discuss these two types of products.
Scalar (or dot ) product of two vector :The scalar product of of two non zero vector a and vector b , denoted by vector \vec{a}\cdot \vec{b} is define as \vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \Theta
where, θ is the angle between and \vec{a}\:and\:\vec{b} , 0 is not defined, and in this case

 

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