Find the values of t for which the velocity of the car is parallel to the vector (i + j)

v = (3r^2 - 2t + 8)i + (5t + 6)j ms^-1

Its doing my head in not being able to work this out (Angry)

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- Sep 26th 2008, 08:09 AMCarl Felthamvectors
Find the values of t for which the velocity of the car is parallel to the vector (i + j)

v = (3r^2 - 2t + 8)i + (5t + 6)j ms^-1

Its doing my head in not being able to work this out (Angry) - Sep 26th 2008, 08:21 AMcivodul
For vectors to be parallel they need to have the same slope.

vector i+j has slope: 1/1= 1

so slope of v has to be 1 which is:

(5t+6)/(3t^2-2t+8)=1

So it is a second degree polynomial equation to solve.

civodul - Sep 26th 2008, 11:41 AMSoroban
Hello, Carl!

Quote:

Find the values of $\displaystyle t$ for which the velocity of the car

is parallel to the vector $\displaystyle \vec i + \vec j$

. . $\displaystyle \vec v \;= \;(3t^2 - 2t + 8)\vec i + (5t + 6)\vec j$

A vector parallel to $\displaystyle \vec i + \vec j$ has the form: .$\displaystyle \vec v \:=\:a\vec i + a\vec j$

. . That is, the coefficients are equal.

So we have: .$\displaystyle 3t^2 - 2t + 8 \;=\;5t + 6 \quad\Rightarrow\quad 3t^2 - 7t + 2 \:=\:0$

Factor: .$\displaystyle (t - 2)(3t - 1) \:=\:0 \quad\Rightarrow\quad\boxed{ t \;=\;2,\:\frac{1}{3}}$

- Sep 26th 2008, 01:56 PMProve It
Another way may be to use dot products.

Recall that

$\displaystyle \mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}$.

Since the vectors will be parallel, the angle between them is 0, and so $\displaystyle \cos{\theta}=\cos{0} = 1$.

So in other words, in this case...

$\displaystyle \mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|$

So evaluate the dot product, evaluate the product of the moduli, and solve for t.