
If vectors \[A = 2{\bf{i}} + 3{\bf{j}} + 4{\bf{k}},\vec B = {\bf{i}} + {\bf{j}} + 5{\bf{k}}\], and \[C\] form a left handed system, then \[C\]is
A. \[11i - 6j - k\]
B. \[ - 11i - 6j + k\]
C. \[11i - 6j + {\rm{k}}\]
D. \[ - 11{\rm{i}} + 6{\rm{j - k}}\]
Answer
164.7k+ views
Hint: The dot product of two vectors \[{\rm{p}}\] and \[{\rm{q}}\] where \[p = xi + yj + zk\] and \[q = {x^\prime }i + {y^\prime }j + {z^\prime }k\] is given by the fact that
\[p \cdot q = x{x^\prime } + y{y^\prime } + z{z^\prime }\]
The cross product of the two vectors will be as follows:
\[p \times q = \left| {\begin{array}{*{20}{c}}i&j&k\\x&y&z\\{{x^\prime }}&{{y^\prime }}&{{z^\prime }}\end{array}} \right|\]
In this case, we have been given two vectors A and B, we have to determine the cross product of the provided to vectors whereas C vector is a left handed vector which is to be find.
Formula Used:The dot product of vectors is given as
\[\vec a \cdot \vec b = |a||b|\cos \theta \]
Scalar product of vectors
\[(p{\bf{a}}).(q{\bf{b}}) = (p{\bf{b}}).(q{\bf{a}}) = pq\;{\bf{a}}.{\bf{b}}\]
Complete step by step solution:We have been given in the question that vectors are
\[A = 2{\bf{i}} + 3{\bf{j}} + 4{\bf{k}},\vec B = {\bf{i}} + {\bf{j}} + 5{\bf{k}}\]
The vector \[\vec C\] are left handed system.
Now, we have to find the \[\vec C\]
For that, we have to write it as
\[\vec C = \vec A \times \vec B\]
Now, we have to write the given vectors in the form of matrix, we obtain
\[ = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\2&3&4\\1&1&5\end{array}} \right|\]
Now, let us solve the matrix, we get
\[ = \hat i(15 - 4) - \hat j(10 - 4) + \hat k(2 - 3)\]
Now, we have to simplify the terms inside the parentheses in the above equation, we get
\[ = \hat i(11) - \hat j(6) + \hat k( - 1)\]
Now, let’s restructure the above equation by forming as follows:
\[ = 11\hat i - 6\hat j - 1\hat k\]
Therefore, if vectors \[A = 2{\bf{i}} + 3{\bf{j}} + 4{\bf{k}},\vec B = {\bf{i}} + {\bf{j}} + 5{\bf{k}}\], and \[C\] form a left handed system, then \[C\]is \[11\hat i - 6\hat j - 1\hat k\]
Option ‘B’ is correct
Note: The dot product and cross product laws should be perfectly remembered and applied. When simplifying, be mindful of the sign and keep in mind the following golden rules: Adding two positive terms results in a positive term. You must deduct from one positive and one negative word to indicate the larger values, whether positive or negative. Even if adding two negative numbers results in a negative number, you actually need to add both values and add a negative sign to the response.
\[p \cdot q = x{x^\prime } + y{y^\prime } + z{z^\prime }\]
The cross product of the two vectors will be as follows:
\[p \times q = \left| {\begin{array}{*{20}{c}}i&j&k\\x&y&z\\{{x^\prime }}&{{y^\prime }}&{{z^\prime }}\end{array}} \right|\]
In this case, we have been given two vectors A and B, we have to determine the cross product of the provided to vectors whereas C vector is a left handed vector which is to be find.
Formula Used:The dot product of vectors is given as
\[\vec a \cdot \vec b = |a||b|\cos \theta \]
Scalar product of vectors
\[(p{\bf{a}}).(q{\bf{b}}) = (p{\bf{b}}).(q{\bf{a}}) = pq\;{\bf{a}}.{\bf{b}}\]
Complete step by step solution:We have been given in the question that vectors are
\[A = 2{\bf{i}} + 3{\bf{j}} + 4{\bf{k}},\vec B = {\bf{i}} + {\bf{j}} + 5{\bf{k}}\]
The vector \[\vec C\] are left handed system.
Now, we have to find the \[\vec C\]
For that, we have to write it as
\[\vec C = \vec A \times \vec B\]
Now, we have to write the given vectors in the form of matrix, we obtain
\[ = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\2&3&4\\1&1&5\end{array}} \right|\]
Now, let us solve the matrix, we get
\[ = \hat i(15 - 4) - \hat j(10 - 4) + \hat k(2 - 3)\]
Now, we have to simplify the terms inside the parentheses in the above equation, we get
\[ = \hat i(11) - \hat j(6) + \hat k( - 1)\]
Now, let’s restructure the above equation by forming as follows:
\[ = 11\hat i - 6\hat j - 1\hat k\]
Therefore, if vectors \[A = 2{\bf{i}} + 3{\bf{j}} + 4{\bf{k}},\vec B = {\bf{i}} + {\bf{j}} + 5{\bf{k}}\], and \[C\] form a left handed system, then \[C\]is \[11\hat i - 6\hat j - 1\hat k\]
Option ‘B’ is correct
Note: The dot product and cross product laws should be perfectly remembered and applied. When simplifying, be mindful of the sign and keep in mind the following golden rules: Adding two positive terms results in a positive term. You must deduct from one positive and one negative word to indicate the larger values, whether positive or negative. Even if adding two negative numbers results in a negative number, you actually need to add both values and add a negative sign to the response.
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