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If the vector $\overrightarrow b = 3\hat j + 4\hat k$ is written as the sum of the vector $\overrightarrow {{b_1}}$, parallel to $\vec a = \hat i + \hat j$ and a vector $\overrightarrow {{b_2}}$, perpendicular to $\vec a$, then $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}}$ is equals to$A.{\text{ }}3\hat i - 3\hat j + 9\hat k \\ B.{\text{ }} - 6\hat i + 6\hat j - \dfrac{9}{2}\hat k \\ C.{\text{ }}6\hat i - 6\hat j + \dfrac{9}{2}\hat k \\ D.{\text{ }} - 3\hat i + 3\hat j - 9\hat k \\$

Last updated date: 20th Jun 2024
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Hint: A quantity having magnitude as well as direction is known as a vector, necessary for finding the position of one point in space with respect to others. Vectors are represented by a line symbol $\to$on the top of the variable that determines the magnitude and direction of the quantity. To check whether the given two vectors are perpendicular, we need to find their scalar, which should be equal to zero, i.e. $\vec a \cdot \vec b = 0$. For two vectors to be parallel, the equation will differ by a common factor only.
In this question, the concept of parallel as well as perpendicular vectors is to be used, such that $\overrightarrow {{b_1}}$ is parallel to a vector and $\overrightarrow {{b_2}}$ is perpendicular to a vector. So, we need to first evaluate the expression for the vectors $\overrightarrow {{b_1}}$ and $\overrightarrow {{b_2}}$ then find their cross product.

Complete step by step solution: A vector $\overrightarrow {{b_1}}$ is parallel to $\vec a = \hat i + \hat j$ so $\overrightarrow {{b_1}}$ can be written as $\overrightarrow {{b_1}} = m\left( {\hat i + \hat j} \right) - - - - (i)$.
Consider the vector $\overrightarrow {{b_2}} = p\hat i + q\hat j + r\hat k$ such that it is perpendicular to $\vec a = \hat i + \hat j$ so, the dot product of $\overrightarrow {{b_2}}$ and the vector $\vec a$ is zero.
Now, according to the question, $\vec b = \overrightarrow {{b_1}} + \overrightarrow {{b_2}}$. So, substitute $\overrightarrow b = 3\hat j + 4\hat k$, $\overrightarrow {{b_1}} = m\left( {\hat i + \hat j} \right)$ and $\overrightarrow {{b_2}} = p\hat i + q\hat j + r\hat k$ to determine the relationship between the coefficients as:
$\vec b = \overrightarrow {{b_1}} + \overrightarrow {{b_2}} \\ 3\hat j + 4\hat k = m(\hat i + \hat j) + \left( {p\hat i + q\hat j + r\hat k} \right) \\ 3\hat j + 4\hat k = \left( {m + p} \right)\hat i + (m + q)\hat j + r\hat k - - - - (iii) \\$
Now, comparing the coefficients in equation (iii), we get:
$3\hat j + 4\hat k = \left( {m + p} \right)\hat i + (m + q)\hat j + r\hat k \\ m + p = 0 \Rightarrow m = - p - - - - (iv) \\ 3 = m + q \Rightarrow m = 3 - q - - - (v) \\ 4 = r - - - - (vi) \\$
Now, by equation (ii), (iv) and (v), we get:
$2m = 3 \\ m = \dfrac{3}{2} \\ p = - \dfrac{3}{2} \\ q = \dfrac{3}{2} \\$
Substitute the value of p, q, and r in the equation $\overrightarrow {{b_1}} = m\left( {\hat i + \hat j} \right)$ and $\overrightarrow {{b_2}} = p\hat i + q\hat j + r\hat k$ to determine the vectors $\overrightarrow {{b_1}}$ and $\overrightarrow {{b_2}}$ as:
$\overrightarrow {{b_1}} = m\left( {\hat i + \hat j} \right) \\ = \dfrac{3}{2}\hat i + \dfrac{3}{2}\hat j \\ \overrightarrow {{b_2}} = p\hat i + q\hat j + r\hat k \\ = - \dfrac{3}{2}\hat i + \dfrac{3}{2}\hat j + 4\hat k \\$
Now, the question is asking for the expression of $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}}$ which can be determined as:
$\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ {\dfrac{3}{2}}&{\dfrac{3}{2}}&0 \\ { - \dfrac{3}{2}}&{\dfrac{3}{2}}&4 \end{array}} \right| \\ = \hat i\left( {\dfrac{3}{2} \times 4 - 0} \right) - \hat j\left( {\dfrac{3}{2} \times 4 - 0} \right) + \hat k\left( {\dfrac{3}{2} \times \dfrac{3}{2} + \dfrac{3}{2} \times \dfrac{3}{2}} \right) \\ = 6\hat i - 6\hat j + \dfrac{9}{2}\hat k \\$
Hence, the $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = 6\hat i - 6\hat j + \dfrac{9}{2}\hat k$
Option C is correct.

Note: It should always be kept in mind while applying the concept of parallel vectors that the cross-product of the two vectors will be zero when they are parallel while the dot-product of the two vectors is zero when they are perpendicular.