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Question: If the three vectors are coplanar, then the value of \('x'\) is: \(\vec A = \hat i - 2\hat j + 3\h...

If the three vectors are coplanar, then the value of ′x′'x' is:
A⃗=i^−2j^+3k^\vec A = \hat i - 2\hat j + 3\hat k, B⃗=xj^+3k^\vec B = x\hat j + 3\hat k and C⃗=7i^+3j^−11k^\vec C = 7\hat i + 3\hat j - 11\hat k
A. 3621\dfrac{{36}}{{21}}
B. −5132 - \dfrac{{51}}{{32}}
C. 5132\dfrac{{51}}{{32}}
D. −3621 - \dfrac{{36}}{{21}}

Explanation

Solution

Here, in the given question, we are given that three vectors (A⃗,B⃗,C⃗)\left( {\vec A,\vec B,\vec C} \right) are coplanar and we need to find the value of ′x′'x'. Three or more vectors lying on the same plane are called coplanar vectors. If three vectors are coplanar then their scalar triple product is zero. So here, we will find the equate the scalar triple product of three vectors equal to zero and then will find the value of xx.

Complete step by step answer:
We have, A⃗=i^−2j^+3k^\vec A = \hat i - 2\hat j + 3\hat k
⇒B⃗=xj^+3k^\Rightarrow \vec B = x\hat j + 3\hat k
⇒C⃗=7i^+3j^−11k^\Rightarrow \vec C = 7\hat i + 3\hat j - 11\hat k
As we know, if three vectors are coplanar, their scalar triple product is equal to zero. So now we will find the scalar product of the three given vectors.
\Rightarrow \left| {\begin{array}{*{20}{c}} 1&{ - 2}&3 \\\ 0&x;&3 \\\ 7&3&{ - 11} \end{array}} \right| = 0

Now, we will expand the above written determinant. We will expand it along the first column.
⇒1(x×−11−3×3)−(−2)(0×−11−3×7)+3(0×3−x×7)=0\Rightarrow 1\left( {x \times - 11 - 3 \times 3} \right) - \left( { - 2} \right)\left( {0 \times - 11 - 3 \times 7} \right) + 3\left( {0 \times 3 - x \times 7} \right) = 0
On solving the terms written inside the brackets, we get
⇒1(−11x−9)+2(−21)+3(−7x)=0\Rightarrow 1\left( { - 11x - 9} \right) + 2\left( { - 21} \right) + 3\left( { - 7x} \right) = 0
⇒−11x−9−42−21x=0\Rightarrow - 11x - 9 - 42 - 21x = 0

On solving the like terms, we get
⇒−32x−51=0\Rightarrow - 32x - 51 = 0
On shifting 5151 to the right hand side, we get
⇒−32x=51\Rightarrow - 32x = 51
∴x=51−32\therefore x = \dfrac{{51}}{{ - 32}}
Thus, if the three given vectors (A⃗,B⃗,C⃗)\left( {\vec A,\vec B,\vec C} \right) are coplanar, then the value of ′x′'x' is 51−32\dfrac{{51}}{{ - 32}}.

Therefore, the correct option is B.

Note: Remember that the conditions for vectors to be coplanar if there are three vectors, is (a) if their scalar triple product is zero, (b) if they are linearly dependent and (c) in case of nn vectors if no more than two vectors are linearly independent. Also, the equation system that has the determinant of the coefficient as zero is called a non-trivial solution. To solve this type of questions, one must know how to multiply terms with opposite signs and same signs. While taking the product of two numbers, always keep in mind that product one positive and one negative number is always negative, product of two negative numbers is positive and product of two positive numbers is positive. One must also know that the sum (or difference) of two like terms is a like term with coefficients equal to the sum (or difference) of coefficients of the two like terms. When we add (or subtract) two algebraic expressions, the like terms are added (or subtracted) and the unlike terms are written as they are.