
The value of \[\lambda \] for which the four points \[2{\bf{i}} + 3{\bf{j}} - {\bf{k}},{\bf{i}} + 2{\bf{j}} + 3{\bf{k}},3{\bf{i}} + 4{\bf{j}} - 2{\bf{k}},{\bf{i}} - \lambda {\bf{j}} + 6{\bf{k}}\] are coplanar
A. \[8\]
B. \[0\]
C. \[ - 2\]
D. \[6\]
Answer
163.2k+ views
Hint: We know that if four points are coplanar, the vectors created by including two of them at a time have a determinant equal to zero. Four points are given to be coplanar (0). The vectors produced by including two of the four coplanar points at a time have a determinant of zero if all four points are coplanar (0
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: We have been given four vectors in the question as follows,
Let \[\vec A = 2\hat i + 3\hat j - \hat k \ldots (1)\]
\[\vec B = \hat i + 2\hat j + 3\hat k\quad \ldots (2)\]
\[\vec C = 3\hat i + 4\hat j - 2\hat k\quad \ldots (3)\]
\[\vec D = \hat i - \lambda \hat j + 6\hat k\quad \ldots (4)\]
Therefore, writing the above vectors in terms of
\[\overrightarrow {AB} = - \hat i - \hat j + 4\hat k\]
\[\overrightarrow {AC} = \hat i + \hat j - \hat k{\rm{ }}\]
\[\overrightarrow {AD} = - \hat i - (\lambda + 3)\hat j + 7\hat k\]
Now, we have to transform the vectors in to the matrix form by using condition of coplanarity, we get
\[\left| {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&4\\1&1&{ - 1}\\{ - 1}&{ - (\lambda + 3)}&7\end{array}} \right| = 0\]
Now, let us applying the relation\[{R_1} \to {R_1} + {R_2}\], we get the resultant in matrix form,
\[\left| {\begin{array}{*{20}{c}}0&0&3\\1&1&{ - 1}\\{ - 1}&{ - (\lambda + 3)}&7\end{array}} \right| = 0\]
Now, we have to solve the matrix along row, we obtain
\[ \Rightarrow - \lambda - 2 = 0\]
On solving the above equation for\[\lambda \], we get
\[ \Rightarrow \lambda = - 2\]
Therefore, the value of \[\lambda \] for which the four points \[2{\bf{i}} + 3{\bf{j}} - {\bf{k}},{\bf{i}} + 2{\bf{j}} + 3{\bf{k}},3{\bf{i}} + 4{\bf{j}} - 2{\bf{k}},{\bf{i}} - \lambda {\bf{j}} + 6{\bf{k}}\] are coplanar is \[\lambda = - 2\]
Option ‘C’ is correct
Note: If the supports of a given set of vectors are parallel to one another, then those vectors are said to be coplanar. Students should keep in mind that two vectors are always in coplanar relationship. Two vectors are also said to be collinear if their supports are parallel, regardless of the direction of the vectors. Keep in mind that parallel vectors are another name for collinear vectors. They are referred to as dissimilar vectors if their directions are opposite one another.
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: We have been given four vectors in the question as follows,
Let \[\vec A = 2\hat i + 3\hat j - \hat k \ldots (1)\]
\[\vec B = \hat i + 2\hat j + 3\hat k\quad \ldots (2)\]
\[\vec C = 3\hat i + 4\hat j - 2\hat k\quad \ldots (3)\]
\[\vec D = \hat i - \lambda \hat j + 6\hat k\quad \ldots (4)\]
Therefore, writing the above vectors in terms of
\[\overrightarrow {AB} = - \hat i - \hat j + 4\hat k\]
\[\overrightarrow {AC} = \hat i + \hat j - \hat k{\rm{ }}\]
\[\overrightarrow {AD} = - \hat i - (\lambda + 3)\hat j + 7\hat k\]
Now, we have to transform the vectors in to the matrix form by using condition of coplanarity, we get
\[\left| {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&4\\1&1&{ - 1}\\{ - 1}&{ - (\lambda + 3)}&7\end{array}} \right| = 0\]
Now, let us applying the relation\[{R_1} \to {R_1} + {R_2}\], we get the resultant in matrix form,
\[\left| {\begin{array}{*{20}{c}}0&0&3\\1&1&{ - 1}\\{ - 1}&{ - (\lambda + 3)}&7\end{array}} \right| = 0\]
Now, we have to solve the matrix along row, we obtain
\[ \Rightarrow - \lambda - 2 = 0\]
On solving the above equation for\[\lambda \], we get
\[ \Rightarrow \lambda = - 2\]
Therefore, the value of \[\lambda \] for which the four points \[2{\bf{i}} + 3{\bf{j}} - {\bf{k}},{\bf{i}} + 2{\bf{j}} + 3{\bf{k}},3{\bf{i}} + 4{\bf{j}} - 2{\bf{k}},{\bf{i}} - \lambda {\bf{j}} + 6{\bf{k}}\] are coplanar is \[\lambda = - 2\]
Option ‘C’ is correct
Note: If the supports of a given set of vectors are parallel to one another, then those vectors are said to be coplanar. Students should keep in mind that two vectors are always in coplanar relationship. Two vectors are also said to be collinear if their supports are parallel, regardless of the direction of the vectors. Keep in mind that parallel vectors are another name for collinear vectors. They are referred to as dissimilar vectors if their directions are opposite one another.
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