
If the vectors \[i + 3j - 2k,2i - j + 4k\] and \[3i + 2j + xk\] are coplanar, then the value of \[{\rm{x}}\] is
A. 2
B. 2
C. 1
D. 3
Answer
162.3k+ views
Hint: Coplanar vectors are those that are located on the same surface of a three-dimensional plane. The same plane is perpendicular to each vector. It is always easy to find any two coplanar random vectors in a plane. In a three-dimensional vector space, the coplanarity between two lines is located. Three vectors are said to be coplanar when their scalar product is zero.
Formula Used:The scalars x, y, and z must not all be zero in order for the vectors a, b, and c to be coplanar.
\[{\bf{xa}} + {\bf{yb}} + {\bf{zc}} = 0\]
Complete step by step solution:We have been given that the vectors\[{\rm{i}} + 3{\rm{j}} - 2{\rm{k}},2{\rm{i}} - {\rm{j}} + 4{\rm{k}}\]and \[3{\rm{i}} + 2{\rm{j}} + {\rm{xk}}\] are coplanar
Now, we have to write the above each vector’s coefficient as determinant, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}1&3&{ - 2}\\2&{ - 1}&4\\3&2&x\end{array}} \right| = 0\]
On solving the determinant from the above, we obtain
\[ \Rightarrow ( - x - 8) - 3(2x - 12) - 14 = 0\]
Now, we have to subtract\[{\rm{( - 8 - 14)}}\]from the above equation, we get
\[ - x - 8 - 3\left( {2x - 12} \right) - 14 - \left( { - 8 - 14} \right) = 0 - \left( { - 8 - 14} \right)\]
On simplifying, we obtain
\[ - x - 3\left( {2x - 12} \right) = 22\]
Now, we have to expand the above equation, we get
\[ - x - 6x + 36 = 22\]
Let’s add similar elements, we get
\[ - 7x + 36 = 22\]
Now, we have to subtract\[{\rm{36}}\]from both sides, we get
\[{\rm{ - 7x + 36 - 36 = 22 - 36}}\]
Again, on simplifying the above equation, we obtain
\[{\rm{ - 7x = - 14}}\]
Now, we have to divide both sides by\[ - 7\]:
\[\frac{{ - 7x}}{{ - 7}} = \frac{{ - 14}}{{ - 7}}\]
Now, let’s simplify to get the resultant value.
\[x = 2\]
Therefore, if the vectors \[i + 3j - 2k,2i - j + 4k\] and \[3i + 2j + xk\] are coplanar, then the value of \[{\rm{x}}\] is \[2\]
Option ‘A’ is correct
Note: Students should be well aware that, three vectors are coplanar if their scalar triple product equals zero in a three-dimensional space. Three vectors in a three-dimensional space are coplanar if they are linearly independent of one another. If three vectors in a three-dimensional space are linearly independent of one another, they are coplanar.
Formula Used:The scalars x, y, and z must not all be zero in order for the vectors a, b, and c to be coplanar.
\[{\bf{xa}} + {\bf{yb}} + {\bf{zc}} = 0\]
Complete step by step solution:We have been given that the vectors\[{\rm{i}} + 3{\rm{j}} - 2{\rm{k}},2{\rm{i}} - {\rm{j}} + 4{\rm{k}}\]and \[3{\rm{i}} + 2{\rm{j}} + {\rm{xk}}\] are coplanar
Now, we have to write the above each vector’s coefficient as determinant, we get
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}1&3&{ - 2}\\2&{ - 1}&4\\3&2&x\end{array}} \right| = 0\]
On solving the determinant from the above, we obtain
\[ \Rightarrow ( - x - 8) - 3(2x - 12) - 14 = 0\]
Now, we have to subtract\[{\rm{( - 8 - 14)}}\]from the above equation, we get
\[ - x - 8 - 3\left( {2x - 12} \right) - 14 - \left( { - 8 - 14} \right) = 0 - \left( { - 8 - 14} \right)\]
On simplifying, we obtain
\[ - x - 3\left( {2x - 12} \right) = 22\]
Now, we have to expand the above equation, we get
\[ - x - 6x + 36 = 22\]
Let’s add similar elements, we get
\[ - 7x + 36 = 22\]
Now, we have to subtract\[{\rm{36}}\]from both sides, we get
\[{\rm{ - 7x + 36 - 36 = 22 - 36}}\]
Again, on simplifying the above equation, we obtain
\[{\rm{ - 7x = - 14}}\]
Now, we have to divide both sides by\[ - 7\]:
\[\frac{{ - 7x}}{{ - 7}} = \frac{{ - 14}}{{ - 7}}\]
Now, let’s simplify to get the resultant value.
\[x = 2\]
Therefore, if the vectors \[i + 3j - 2k,2i - j + 4k\] and \[3i + 2j + xk\] are coplanar, then the value of \[{\rm{x}}\] is \[2\]
Option ‘A’ is correct
Note: Students should be well aware that, three vectors are coplanar if their scalar triple product equals zero in a three-dimensional space. Three vectors in a three-dimensional space are coplanar if they are linearly independent of one another. If three vectors in a three-dimensional space are linearly independent of one another, they are coplanar.
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