Question

Find the magnitude of component of 3i^ - 2j^ +k^ along the vector 12i^+ 3j^ - 4k^. (^ = cap)

Answer 1

not confirm about answer.......
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Answer 2

Answer:

The magnitude of component of vector A along the vector B is 2units.

Explanation:

Given two vectors, $\vec A=3\hat i - 2\hat j +\hat k$ and $\vec B=12\hat i+3\hat j-4\hat k$

Need to find the component of vector A along vector B.

This refers to the projection of one vector onto the other vector.

• The vector projection of vector A on vector B is given by

        $proj_uv=\dfrac{u.v}{|u|}\dfrac{u}{|u|}$

• where ${u}/{|u|}$ is the unit vector and ${u.v}/{|u|}$ is the magnitude of the vector projection called the scalar projection .
• A scalar projection is given by the dot product of a vector with a unit vector for that direction.

Thus, applying the scalar projection of vector A on vector B, given by

$\dfrac{\vec A.\vec B}{|\vec B|}=\dfrac{(3\hat i - 2\hat j +\hat k)(12\hat i+3\hat j-4\hat k)}{\sqrt{(12)^{2}+3^{2} +(-4)^{2} } }$

$=\dfrac{(3\times 12)\hat i .\hat i +(- 2\times3)\hat j.\hat j +(1\times(-4)\hat k.\hat k }{\sqrt{144+9 +16} }$

since $\hat i .\hat i =\hat j.\hat j =\hat k.\hat k =1$, we get

$\dfrac{\vec A.\vec B}{|\vec B|}=\dfrac{36- 6 -4}{\sqrt{169} }=\dfrac{26}{13 }=2$

Therefore, the magnitude of component of vector A along the vector B is 2units.

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