Question

How do you find the projection of $u$ onto $v$ given $u = < 3,15 >$ and $< v = - 1,5 >$ ?

Answer 1

To solve this question first we have to use the formula of the dot product. We simplify the formula in terms of projection and then we use dot product and find the modulus of vector $v$ and then put the value in the formula that we simplified that earlier and that is the projection.

Complete step by step answer:
We have given two vectors.
$u = < 3,15 >$ and $v = < \- 1,5 >$
To find the projection of $u$ onto $v$ we use the dot product of two vectors.
$\vec u.\vec v = \left\| {\vec u} \right\|\left\| {\vec v} \right\|\cos \theta$
The projection on $u$ onto $v$ is $\left\| {\vec u} \right\|\cos \theta$ so on, simplifying this formula in terms of projection.
$\left\| {\vec u} \right\|\cos \theta = \dfrac{{\vec u.\vec v}}{{\left\| {\vec v} \right\|}}$

Now we have to find the value of the dot product of both the vectors. In dot product x coordinate is multiplied with x coordinate of another vector and y coordinate of one vector to y coordinate of another vector.
$\vec u.\vec v = \left\langle {3,15} \right\rangle .\left\langle { - 1,5} \right\rangle$
On further solving
$\vec u.\vec v = - 3 + 75$
On further calculating
$\vec u.\vec v = 72$

Now we have to find the value of the modulus of the vector $\vec v$. Modulus is the magnitude of the vector and that is the same as the distance formula between two points.
$\left\| {\vec v} \right\| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 5 \right)}^2}}$
On further solving
$\left\| {\vec v} \right\| = \sqrt {1 + 25}$
On adding both the numbers.
$\left\| {\vec v} \right\| = \sqrt {26}$
On putting all these values in the formula.
$\therefore \left\| {\vec u} \right\|\cos \theta = \dfrac{{72}}{{\sqrt {26} }}$

Therefore, the projection of $u$ onto $v$ is $\left\| {\vec u} \right\|\cos \theta = \dfrac{{72}}{{\sqrt {26} }}$.

Note: To solve these types of questions students must know all the formulas of vectors and concepts of vectors. There are many places where students often make mistakes like exchanging the value of both vectors. The vectors are given in different forms so we just make it like the standard of just use as it is.