# Find the vector that has the same direction as (6, 2, -3) but has length 4.

Find the vector that has the same direction as (6, 2, -3) but has length 4.
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averes8
Princepies: Two vectors  and $<{x}_{2},{y}_{2},{z}_{2}>$ is said to be parallel “have the same direction”, when they are multiple of each other such that
$$=n$$ (1) For example, vectors < 1,2,3 > and < 2,4,6> are said to be parallel and have the same direction as they are bath multiples of each other the second vector is twice the first vector.
And the length of the vector %=< a,b,c > “the magnitude” is given by the following formula
$\stackrel{\to }{u}=\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}$ (2)
And the formula for the dot product between two vectors vec a and vec b is given by
$\stackrel{\to }{a}\cdot \stackrel{\to }{b}=|a||b|\mathrm{cos}\left(\theta \right)$
And, since the problem is asking for what a vector having the same direction but a different length, thus it is asking for a different multiple for the original vector as we are goine to see in the follaving sections.
Givens:
It is given that the direction of the vector is < 6,2,—3 >, and it is asking for another vector having the same direction but with a different length. Since the vector having the same direction there for, it is a multiple of the original vector such that it is direction using equation (1) is given by
n<6,2,-3> (4)
Where, n can be any real number. And since it is asking for it to have a length of 4 units, there for the length of the vector <6n,2n,—3n > must be equal to 4.
Calculations:
Given that the required vector should have the same direction and from the given (4), we know that the direction of the required vector is
n<6,2,-3>
And, since it is length is equal 4, we can find n thus we can find therequired vector as follows. The length of the vector, using equation (2) is
$\sqrt{36{n}^{2}+4{n}^{2}+9{n}^{2}}=4$
$\sqrt{36+4+9}=4$
$n\cdot \sqrt{4}9=4$
$7n=4$
Thus, the multiple n is
$n=4/7$
And hence, the direction of the required vector is
$<6n,2n,-3n>=<\frac{24}{7},\frac{8}{7},\frac{-12}{7}>$
Checking our answer we find that the length of the vector <24/7,8/7,-12/7> using equation (2) is
$\sqrt{\left(\frac{24}{7}{\right)}^{2}+\left(\frac{8}{7}{\right)}^{2}+\left(\frac{-12}{7}{\right)}^{2}}=\sqrt{1}6=4$
And, using the dot product between the two vectors < 6,2,—3 > and <24/7,8/7,-12/7> to find whether they are parallel or not we find that there dot fuct is equal to 28, Where
$<6,2,-3>\cdot <\frac{24}{7},\frac{8}{7},\frac{-12}{7}>=\frac{6\cdot 24}{7}+\frac{2\cdot 8}{7}+\frac{-3\cdot \left(-12\right)}{7}=28$
And using the formula for the dot product (3), and since the dot product is 28 and the magnitude of the first vector is 7 and the second vector is 4, we get
$\mathrm{cos}\left(\theta \right)=\frac{28}{4\cdot 7}=1$
Thus, the angle between both vectors is zero thus they are both parallel, there for the chisined vector represent the reaquired vector.
Results
The required vector is $<\frac{24}{7},\frac{8}{7},\frac{-12}{7}>.$