Question
Question: If the vector \(\overrightarrow b = 3\hat j + 4\hat k\) is written as the sum of the vector \(\overr...
If the vector b=3j^+4k^ is written as the sum of the vector b1, parallel to a=i^+j^ and a vector b2, perpendicular to a, then b1×b2 is equals to
A. 3i^−3j^+9k^ B. −6i^+6j^−29k^ C. 6i^−6j^+29k^ D. −3i^+3j^−9k^
Solution
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 →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. a⋅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 b1 is parallel to a vector and b2 is perpendicular to a vector. So, we need to first evaluate the expression for the vectors b1 and b2 then find their cross product.
Complete step by step solution: A vector b1 is parallel to a=i^+j^ so b1 can be written as b1=m(i^+j^)−−−−(i).
Consider the vector b2=pi^+qj^+rk^ such that it is perpendicular to a=i^+j^ so, the dot product of b2 and the vector a is zero.
Now, according to the question, b=b1+b2. So, substitute b=3j^+4k^, b1=m(i^+j^) and b2=pi^+qj^+rk^ to determine the relationship between the coefficients as:
b=b1+b2 3j^+4k^=m(i^+j^)+(pi^+qj^+rk^) 3j^+4k^=(m+p)i^+(m+q)j^+rk^−−−−(iii)
Now, comparing the coefficients in equation (iii), we get:
3j^+4k^=(m+p)i^+(m+q)j^+rk^ m+p=0⇒m=−p−−−−(iv) 3=m+q⇒m=3−q−−−(v) 4=r−−−−(vi)
Now, by equation (ii), (iv) and (v), we get:
2m=3 m=23 p=−23 q=23
Substitute the value of p, q, and r in the equation b1=m(i^+j^) and b2=pi^+qj^+rk^ to determine the vectors b1 and b2 as:
Now, the question is asking for the expression of b1×b2 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 b1×b2=6i^−6j^+29k^
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.