
Vector coplanar with vectors \[{\rm{i}} + {\rm{j}}\]and \[{\rm{j}} + {\rm{k}}\] parallel to the vector \[2{\rm{i - }}2{\rm{j - }}4{\rm{k}}\] is
A. \[i - k\]
B. \[i - j - 2k\]
C. \[i + j - k\]
D. \[3i + 3j - 6k\]
Answer
164.1k+ views
Hint: A linear combination of the other two vectors can be used to express any vector. These are the conditions if the vectors are coplanar and if three vectors are coplanar. If they are existing in 0 space, their scalar product is zero. The vectors are linearly independent in three dimensions. All vectors are coplanar if there are more than two that are linearly independent. In this case, we have to write the given vectors in matrix form with respect to the vector and solve the matrix to obtain the parallel vector.
Formula Used: Vectors \[\vec a,\vec b,\vec c\] are coplanar if their scalar triple product is zero.
\[[\vec a\vec b\vec c] = 0\]
Complete step by step solution: Let us consider the vector is
\[a\hat i + b\hat j + c\hat k\]
We have been already known that\[a\hat i + b\hat j + c\hat k,\hat i + \hat j\]\[\hat j + \hat k\]are coplanar.
And, the matrix form of the vector will be a follows:
\[\left| {\begin{array}{*{20}{l}}a&b&c\\1&1&0\\0&1&1\end{array}} \right| = 0\]
On solving the matrix we will obtain,
\[ \Rightarrow a - b + c = 0\]
Also from the given data it is understand that, \[(a\hat i + b\hat j + c\hat k)\]is parallel to \[(2\hat i - 2\hat j - 4\hat k)\]
Therefore, we can write the equation for the above statement as,
\[(a\hat i + b\hat j + c\hat k) \times (2\hat i - 2\hat j - 4\hat k) = \vec 0\]
Now, let us solve write the equation obtained above in matrix format, we obtain
\[\left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\a&b&c\\2&{ - 2}&{ - 4}\end{array}} \right| = 0\]
Now, we have to solve the matrix along the row, we get\[ \Rightarrow \hat i( - 4b + 2c) - \hat j( - 4a - 2c) + \hat k( - 2a - 2b) = 0\]
Now, let us equate each term in the above equation to zero, we get\[ \Rightarrow - 4b + 2c = 0,4a + 2c = 0,2a + 2b = 0\]
Now, the above equation can be written as,
\[\dfrac{a}{{ - 1}} = \dfrac{b}{1} = \dfrac{c}{2}\]
Or it can also be written as,
\[\dfrac{a}{1} = \dfrac{b}{{ - 1}} = \dfrac{c}{{ - 2}}\]
Thus, from the obtained expression, we will obtain the vectors as
\[\hat i - \hat j - 2\hat k\]
Therefore, vector coplanar with vectors \[{\rm{i}} + {\rm{j}}\]and \[{\rm{j}} + {\rm{k}}\] parallel to the vector \[2{\rm{i - }}2{\rm{j - }}4{\rm{k}}\] is \[\hat i - \hat j - 2\hat k\]
Option ‘B’ is correct
Note: Students are most likely make mistakes while solving with vectors, first they should know vectors and their concepts. A vector is a linear combination of vectors \[{v_1},....{v_n}\] with coefficients\[{a_1},....{a_n}\], where \[{a_1}{v_1} + ........{a_n}{v_n}\]a linear combination of the form \[{a_1}{v_1} + ........{a_n}{v_n}\] is referred to as trivial if all of its coefficients are zero, and it is referred to as non-trivial if at least one of its coefficients is not zero. When two lines in a three-dimensional space are located on the same plane, they are said to be coplanar.
Formula Used: Vectors \[\vec a,\vec b,\vec c\] are coplanar if their scalar triple product is zero.
\[[\vec a\vec b\vec c] = 0\]
Complete step by step solution: Let us consider the vector is
\[a\hat i + b\hat j + c\hat k\]
We have been already known that\[a\hat i + b\hat j + c\hat k,\hat i + \hat j\]\[\hat j + \hat k\]are coplanar.
And, the matrix form of the vector will be a follows:
\[\left| {\begin{array}{*{20}{l}}a&b&c\\1&1&0\\0&1&1\end{array}} \right| = 0\]
On solving the matrix we will obtain,
\[ \Rightarrow a - b + c = 0\]
Also from the given data it is understand that, \[(a\hat i + b\hat j + c\hat k)\]is parallel to \[(2\hat i - 2\hat j - 4\hat k)\]
Therefore, we can write the equation for the above statement as,
\[(a\hat i + b\hat j + c\hat k) \times (2\hat i - 2\hat j - 4\hat k) = \vec 0\]
Now, let us solve write the equation obtained above in matrix format, we obtain
\[\left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\a&b&c\\2&{ - 2}&{ - 4}\end{array}} \right| = 0\]
Now, we have to solve the matrix along the row, we get\[ \Rightarrow \hat i( - 4b + 2c) - \hat j( - 4a - 2c) + \hat k( - 2a - 2b) = 0\]
Now, let us equate each term in the above equation to zero, we get\[ \Rightarrow - 4b + 2c = 0,4a + 2c = 0,2a + 2b = 0\]
Now, the above equation can be written as,
\[\dfrac{a}{{ - 1}} = \dfrac{b}{1} = \dfrac{c}{2}\]
Or it can also be written as,
\[\dfrac{a}{1} = \dfrac{b}{{ - 1}} = \dfrac{c}{{ - 2}}\]
Thus, from the obtained expression, we will obtain the vectors as
\[\hat i - \hat j - 2\hat k\]
Therefore, vector coplanar with vectors \[{\rm{i}} + {\rm{j}}\]and \[{\rm{j}} + {\rm{k}}\] parallel to the vector \[2{\rm{i - }}2{\rm{j - }}4{\rm{k}}\] is \[\hat i - \hat j - 2\hat k\]
Option ‘B’ is correct
Note: Students are most likely make mistakes while solving with vectors, first they should know vectors and their concepts. A vector is a linear combination of vectors \[{v_1},....{v_n}\] with coefficients\[{a_1},....{a_n}\], where \[{a_1}{v_1} + ........{a_n}{v_n}\]a linear combination of the form \[{a_1}{v_1} + ........{a_n}{v_n}\] is referred to as trivial if all of its coefficients are zero, and it is referred to as non-trivial if at least one of its coefficients is not zero. When two lines in a three-dimensional space are located on the same plane, they are said to be coplanar.
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