Question

Find a vector of magnitude 3 in the direction of v = 10i - 24k. The vector is Oi-Oj+k (Simplify your answer. Use integers or fractions for any numbers in the expression).

Answer 1

Given :

A vector v = 10i - 24k.

To Find :

A vector of magnitude 3 in the direction of v = 10i - 24k.

Solution :

Unit vector in the direction  of vector v is :

$\^{v}=\dfrac{v}{|v|}\\\\\^{v}=\dfrac{10i-24k}{\sqrt{10^2+24^2}}\\\\\^{v}=\dfrac{10i-24k}{26}$

Now , vector of magnitude 3 in direction v is :

$r=3 \^{v}\\\\r=3\times \dfrac{ (10i-24k)}{26}\\\\r=\dfrac{3}{13} (5i-12k)$

Hence , this is the required solution .

Answer 2

The vector of magnitude 3 in the direction of given vector is, $\frac{30i-72k}{26}$

Unit vector :

First we have to calculate unit vector of given vector.

    Given vector,  $v=10i-24k$

Unit vector of vector v is,

                     $v=\frac{10i-24k}{\sqrt{(10)^{2}+(24)^{2} } }\\ \\v=\frac{10i-24k}{\sqrt{100+576} }=\frac{10i-24k}{26}$

For vector have magnitude 3 is,

        $v=3(\frac{10i-24k}{26} )=\frac{30i-72k}{26}$

The vector of magnitude 3 in the direction of given vector is, $\frac{30i-72k}{26}$

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