Question

If vector A = 6i - 8j then 4A magnitude has

Answer 1

QUESTION:

The correct question should be:

If vector A = 6i - 8j then vector 4A has a magnitude ____?

ANSWER:

For the given vector A, the magnitude of vector 4A will be 40 units.

Given,

Vector A = 6i - 8j.

To find,

The magnitude of vector 4A.

Solution,

Here, the given vector is,

$\vec A=6i-8j.$

Let there be a vector P, such that,

$\vec P=xi+yj$

Then the magnitude of vector P is given by

$|\vec P|=\sqrt{x^{2} +y^{2}}$

For the given vector A, we have to find the magnitude of vector 4A, so first, we should find the vector 4A by multiplying the given vector A by 4.

This can be done as follows.

$\because \vec A=6i-8j$

$\therefore \vec {4A}=4(6i-8j)$

$\implies \vec {4A}=24i-32j$

Now, the magnitude of 4A will be,

$|\vec {4A}|=\sqrt{(24)^2+(-32)^2}$

$\implies |\vec {4A}|=\sqrt{(576)+(1024)}$

$\implies |\vec {4A}|=\sqrt{1600}$

$\implies |\vec {4A}|=40 \hspace{3} units.$

Therefore, for the given vector A, the magnitude of vector 4A will be 40 units.

Answer 2

Answer:

The Magnitude=40 units

Explanation:

granted,

Vector A = 6i - 8j.

Fine,

Vector size 4A.

Solution,

Here is the given vector,

A=6i-8j

Let there exist a vector P, so

P=xi+yj

Then the size of the vector P is given by the relation

$P=\sqrt{x^{2} +y^{2} }$

For a given vector A we need to find the size of the vector 4A, so first we have to find the vector 4A by multiplying the given vector A by 4.

It can be as follows.

A=6i-8j

4A=4(6i-8j)

4A=24i-32j

Now the size can be 4A,

4A=$\sqrt{} 24^{2} +(-32)^{2}$

=$\sqrt576+1024$

==40

Therefore, for a given vector A, the size of vector 4A would be 40 units.

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