If the vector $$\vec{b} = 3\hat{j} + 4\hat{k}$$ is written as the sum of a vector $$\vec{b_1}$$, parallel to $$\vec{a} = \hat{i} + \hat{j}$$ and a vector $$\vec{b_2}$$, perpendicular to $$\vec{a}$$, then $$\vec{b_1} \times \vec{b_2}$$ is equal to:

We first write down the two given vectors in component form

$$\vec{a}=1\,\hat{i}+1\,\hat{j}+0\,\hat{k}\;,-(1)$$

$$\vec{b}=0\,\hat{i}+3\,\hat{j}+4\,\hat{k}\;.-(2)$$

Let the part of $$\vec{b}$$ that is parallel to $$\vec{a}$$ be $$\vec{b_1}$$. Any vector parallel to $$\vec{a}$$ must be some real multiple of $$\vec{a}$$, so we set

$$\vec{b_1}=t\,\vec{a}\;,-(3)$$

where $$t$$ is a scalar to be determined. The remaining part

$$\vec{b_2}=\vec{b}-\vec{b_1}\;,-(4)$$

is by construction perpendicular to $$\vec{a}$$, i.e.

$$\vec{a}\cdot\vec{b_2}=0\;.-(5)$$

Substituting $$\vec{b_2}$$ from -(4) into the perpendicularity condition -(5) gives

$$\vec{a}\cdot(\vec{b}-\vec{b_1})=0\;.-(6)$$

Expand the dot product in -(6):

$$\vec{a}\cdot\vec{b}-\vec{a}\cdot\vec{b_1}=0\;.-(7)$$

Compute each term separately. First, using -(1) and -(2),

$$\vec{a}\cdot\vec{b}=(1)(0)+(1)(3)+(0)(4)=3\;.-(8)$$

For the second dot product we use -(3) and the distributive law:

$$\vec{a}\cdot\vec{b_1}=\vec{a}\cdot(t\vec{a})=t(\vec{a}\cdot\vec{a})=t|\vec{a}|^{2}\;.-(9)$$

The squared magnitude of $$\vec{a}$$ comes directly from -(1):

$$|\vec{a}|^{2}=(1)^{2}+(1)^{2}+(0)^{2}=2\;.-(10)$$

Insert -(8), -(9) and -(10) into equation -(7):

$$3-t(2)=0\;,-(11)$$

from which

$$t=\dfrac{3}{2}\;.-(12)$$

Now use -(3) and -(12) to obtain $$\vec{b_1}$$ explicitly:

$$\vec{b_1}=\dfrac{3}{2}(1\,\hat{i}+1\,\hat{j}+0\,\hat{k})=\dfrac{3}{2}\hat{i}+\dfrac{3}{2}\hat{j}+0\,\hat{k}\;.-(13)$$

Next, calculate $$\vec{b_2}$$ by substituting -(13) into -(4):

$$\vec{b_2}=(0\,\hat{i}+3\,\hat{j}+4\,\hat{k})-\left(\dfrac{3}{2}\hat{i}+\dfrac{3}{2}\hat{j}+0\,\hat{k}\right)\;,-(14)$$

and subtract component-wise:

$$\vec{b_2}=\Bigl(0-\dfrac{3}{2}\Bigr)\hat{i}+\Bigl(3-\dfrac{3}{2}\Bigr)\hat{j}+(4-0)\hat{k}=-\dfrac{3}{2}\hat{i}+\dfrac{3}{2}\hat{j}+4\hat{k}\;.-(15)$$

As a quick check, verify the perpendicularity using -(1) and -(15):

$$\vec{a}\cdot\vec{b_2}=1\Bigl(-\dfrac{3}{2}\Bigr)+1\Bigl(\dfrac{3}{2}\Bigr)+0(4)=0,$$

so the condition is satisfied.

The desired quantity is the cross product $$\vec{b_1}\times\vec{b_2}$$. Using -(13) and -(15) we write the 3×3 determinant form:

$$ \vec{b_1}\times\vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \dfrac{3}{2} & \dfrac{3}{2} & 0 \\ -\dfrac{3}{2} & \dfrac{3}{2} & 4 \end{vmatrix}\;.-(16) $$

Now expand the determinant component by component.

• $$\hat{i}$$-component:

$$ \left(\dfrac{3}{2}\right)(4)-0\left(\dfrac{3}{2}\right)=6\;.-(17) $$

• $$\hat{j}$$-component (remember the negative sign):

$$ -\Bigl[\left(\dfrac{3}{2}\right)(4)-0\Bigl(-\dfrac{3}{2}\Bigr)\Bigr]=-6\;.-(18) $$

• $$\hat{k}$$-component:

$$ \left(\dfrac{3}{2}\right)\left(\dfrac{3}{2}\right)-\left(\dfrac{3}{2}\right)\Bigl(-\dfrac{3}{2}\Bigr)=\dfrac{9}{4}+\dfrac{9}{4}=\dfrac{9}{2}\;.-(19) $$

Collecting the three components from -(17), -(18) and -(19) gives

$$\vec{b_1}\times\vec{b_2}=6\,\hat{i}-6\,\hat{j}+\dfrac{9}{2}\,\hat{k}\;.-(20)$$

This matches exactly the vector in Option A.

Hence, the correct answer is Option A.