Question

A = 2 î+ root7 j and B =5î -√7j -3k then find the vector whose magnitude is equal to A.B and Parallel to B-A​

Answer 1

Answer:

Let `bar a=2hati-qhatj+3hatk and barb=4hati-5hatj+6hatk`

Since, `bara and barb` are collinear.

∴ there exists a scalar t such that `barb = t bara `.

∴ `4hati-5hatj+6hatk=t(2hati-qhatj+3hatk)=2thati-qthatj+3thatk`

∴ By equality of vectors, we get

4 = 2t, - 5 = - qt, 6 = 3t

∵ 4 = 2t and 6 = 3t ∴t = 2

- 5 =- q(2)

–5 = – 2q

∴ 5 = 2q

q = 5/2

Answer 2

Answer:

The vector whose magnitude is equal to $\vec A.\vec B$ and parallel to $|\vec B-\vec A|$ is $\dfrac{3}{\sqrt{46}}\hat i-\dfrac{2\sqrt{7}}{\sqrt{46}}\hat j-\dfrac{3}{\sqrt{46}}\hat k$    

Explanation:

Given the vectors, $\vec A = 2 \hat i+ \sqrt{7} \hat j$ and $\vec B = 5 \hat i-\sqrt{7} \hat j-3\hat k$

Product of two vectors $\vec A.\vec B$ is

$\vec A.\vec B = (2 \hat i+ \sqrt{7} \hat j) (5 \hat i-\sqrt{7} \hat j-3\hat k)$

       $= (2\times5)+ (\sqrt{7} \times-\sqrt{7})$    

       $= 10- 7=3units$

Therefore, the magnitude of  $\vec A.\vec B$ is 3units.

Then the difference between the vectors, $\vec B-\vec A$ is

$\vec B-\vec A =(5 \hat i-\sqrt{7} \hat j-3\hat k)- (2 \hat i+ \sqrt{7} \hat j)$

          $= (5 \hat i-2 \hat i )+(-\sqrt{7} \hat j-\sqrt{7} \hat j)-3\hat k$

          $=3 \hat i -2\sqrt{7}\hat j -3\hat k$

And the magnitude of the vector $|\vec B-\vec A|$ is

$\big|3 \hat i -2\sqrt{7}\hat j -3\hat k\big|=\sqrt{3^{2}+ (-2\sqrt{7})^{2}+(-3)^{2}}$

                          $=\sqrt{9+28+9}=\sqrt{46}units$

Given the vector that is parallel to $3 \hat i -2\sqrt{7}\hat j -3\hat k$ is equal to 3 units.

A unit vector parallel to another vector is given by  

$\hat a=\dfrac{ \vec a }{|\vec a|}$

Thus, the vector parallel to $3 \hat i -2\sqrt{7}\hat j -3\hat k$ whose magnitude is 3 units is given by

$3\hat a=\dfrac{3 \hat i -2\sqrt{7}\hat j -3\hat k}{|3 \hat i -2\sqrt{7}\hat j -3\hat k|}$

    $=\dfrac{3 \hat i -2\sqrt{7}\hat j -3\hat k}{\sqrt{46}}$      

    $=\dfrac{3}{\sqrt{46}}\hat i-\dfrac{2\sqrt{7}}{\sqrt{46}}\hat j-\dfrac{3}{\sqrt{46}}\hat k$

Therefore, the vector whose magnitude is equal to $\vec A.\vec B$ and parallel to $|\vec B-\vec A|$ is $\dfrac{3}{\sqrt{46}}\hat i-\dfrac{2\sqrt{7}}{\sqrt{46}}\hat j-\dfrac{3}{\sqrt{46}}\hat k$    

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