Question

If a=2i+2j+k, a.b=14, axb=3i+j-8k,
then b =
1) 5i-j+2k 2) 5i+j-k
3) 5i+j+2k 4) 5i+j-2k​

Answer 1

Answer:

$\large\boxed{\sf{(3)\;5\hat{ i } + \hat{j}+2\hat{k}}}$

Step-by-step explanation:

From the question, given that,

$\vec{ a } = 2\hat{ i} + 2\hat{j } + \hat{ k } \\ \\ \vec{ a }.\vec{ b} = 14 \\ \\ \vec{ a } \times \vec{ b} = 3\hat{ i } + \hat{ j } - 8\hat{ k }$

To find the $\vec{b}$, let's assume that,

$\vec{ b } = x\hat{ i } + y\hat{ j } +z \hat{ k }$

Therefore, we will get the dot product as,

$= > \vec{ a }.\vec{ b } = 2x + 2y + z \\ \\ = > 2x + 2y + z = 14 \: \: \: \: \: \: \:............ (1)$

Also, finding the cross product, we get,

$= > \vec{ a } \times \vec{ b} = (2\hat{ i } + 2\hat{ j } + \hat{ k }) \times (x\hat{ i } + y\hat{ j} + z\hat{ k})$

But, we know that,

• $\hat{ i } \times \hat{ i} = \hat{ j } \times \hat{ j } = \hat{ k } \times \hat{ k } = 0$
• $\hat{ i } \times \hat{ j} = \hat{ k }$
• $\hat{ j } \times \hat{ k } = \hat{ i }$
• $\hat{k } \times \hat{ i } = \hat{ j }$

Therefore, we will get,

$= > \vec{ a} \times \vec{b } = (2z - y)\hat{ i } + (x - 2z)\hat{ j } + (2 - 2x)\hat{k } \\ \\ = > (2z - y)\hat{ i } + (x - 2z)\hat{ j } + (2y - 2x)\hat{k } = 3\hat{ i } + \hat{ j } - 8\hat{ k }$

Comparing the coefficient, we have,

$2z - y = 3 \: \: \: \: \: .......(2) \\ \\ x - 2z = 1 \: \: \: \: \: \: \: \: \: .............(3)\\ \\ 2y - 2x = - 8 \: \: \: \: \: \: ..........(4)$

From eqn (2)and (3), we have,

$= > x = 1 + 2z \\ \\ = > y = 2z - 3$

Now, substituting these values in eqn (1), we get,

$= > 2(1 + 2z) + 2(2z - 3) + z = 14 \\ \\ = > 2 + 4z + 4z - 6 + z = 14 \\ \\ = > 9z - 4 = 14 \\ \\ = > 9z = 14 + 4 = 18 \\ \\ = > z = \frac{18}{9} \\ \\ = > z= 2$

Therefore, we will get,

$= > x = 1 +( 2 \times 2) = 1 + 4 \\ \\ = > x = 5$

and,

$= > y =( 2 \times 2) - 3 = 4 - 3\\ \\ = > y = 1$

Hence, we get the required vector,

$\bold{(3) \: \vec{ b } = 5\hat{ i } + \hat{ j } + 2\hat{ k }}$