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If the vector $\underset{b}{\rightarrow}$ $= 3\hat{j}+4\hat{k}$ is written as the sum of a vector $\underset{b_{1}}{\rightarrow}$ , parallel to $\underset{a}{\rightarrow}$ $= \hat{i}+\hat{j}$ , and a vector $\underset{b_{2}}{\rightarrow}$ perpendicular to

$\underset{a}{\rightarrow}$ then $\underset{b_{1}}{\rightarrow}\times \underset{b_{2}}{\rightarrow}$ is equal to :

Option 1)
$-3\hat{i}+3\hat{j}-9\hat{k}$

Option 2)
$6\hat{i}-6\hat{j}+\frac{9}{2}\hat{k}$

Option 3)
$-6\hat{i}+6\hat{j}-\frac{9}{2}\hat{k}$

Option 4)
$3\hat{i}-3\hat{j}+9\hat{k}$

Answers (1)

best_answer

As we have learned

Collinear Vectors -

Two vectors are said to be collinear if and only if there exists a scalar m such as that $\vec{a}=m\vec{b}$

- wherein

m is a Scalar.

Properties of Scalar Product -

$\vec{a}.\vec{b}=0$

$\vec{a}\:is\: prependicular\:\vec{b}$

- wherein

Provided that $\vec{a} \neq 0, \:\vec{b}\neq 0$

Vector Product of two vectors -

Vector product

- wherein

$\vec{a}= (a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k})$

$\vec{b}= (b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k})$

$3 \hat j + 4 \hat k = \vec b_1 + \vec b_2 \\ \vec b_1 = \lambda (\hat i + \hat j )$

$\vec b_2 \cdot \vec a = 0$

$(x\hat i + y \hat j + z\hat k )(\hat i + \hat j ) = 0$

$x+y = 0 \Rightarrow x = -y$

So

$3 \hat j + 4 \hat k = \lambda (\hat i + \hat j )+ (x \hat i - x + z \hat k )$

$0 = \lambda + x = 0 ; \lambda -x = 3 ; z = 4$

$\lambda = 3/2$

$x = -3/2 \\ z = 4$

$\vec b _1 \times vec b_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3/2 & 3/2&0 \\ -3/2 & 3/2 &4 \end{vmatrix}$

= $6 \hat i -6 \hat j + 9/2 \hat k$

Option 1)

$-3\hat{i}+3\hat{j}-9\hat{k}$

Option 2)

$6\hat{i}-6\hat{j}+\frac{9}{2}\hat{k}$

Option 3)

$-6\hat{i}+6\hat{j}-\frac{9}{2}\hat{k}$

Option 4)

$3\hat{i}-3\hat{j}+9\hat{k}$

Posted by

Himanshu

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