
The vectors \[2\widehat{i}+3\widehat{j}-4\widehat{k}\ and\ a\widehat{i}+b\widehat{j}+c\widehat{k}\] are perpendicular, when
A) a=2, b=3, c=-4
B) a=4, b=4, c=5
C) a=4, b=4, c=-5
D) None of these
Answer
614.4k+ views
HINT: - Now, the fact that is important to know before answering this question is that the dot product of two vectors which are perpendicular is equal to 0 that can be written as follows
\[\overrightarrow{v}\cdot \overrightarrow{u}=0\]
(Where \[\overrightarrow{v}\ and\ \overrightarrow{u}\] are two perpendicular vectors)
Another thing is that to calculate a dot product, we do the following
Let \[\overrightarrow{v}=a\widehat{i}+b\widehat{j}+c\widehat{k}\ and\ \overrightarrow{u}=l\widehat{i}+m\widehat{j}+n\widehat{k}\] be two vectors of which we have to calculate the dot product.
Now, to calculate their dot product, we can write the following \[\]
\[\begin{align}
& \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)\ \cdot \left( l\widehat{i}+m\widehat{j}+n\widehat{k} \right) \\
& \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=al+bm+cn \\
\end{align}\]
Complete step-by-step answer:
As mentioned in the question, vectors \[2\widehat{i}+3\widehat{j}-4\widehat{k}\ and\ a\widehat{i}+b\widehat{j}+c\widehat{k}\] are perpendicular to each other.
Now, as we know that the dot product of two vectors which are perpendicular is equal to 0, hence, we can find the required values of a, b and c as follows
\[\begin{align}
& \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)\ \cdot \left( l\widehat{i}+m\widehat{j}+n\widehat{k} \right)=al+bm+cn \\
& \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=\left( 2\widehat{i}+3\widehat{j}-4\widehat{k} \right)\ \cdot \ \left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)=2a+3b-4c \\
\end{align}\]
Now, the dot product must be zero, so we can substitute the values of a, b and c from the options and then we can get the right combination.
Now, for (A), we get
\[\begin{align}
& \Rightarrow 2a+3b-4c \\
& \Rightarrow 2\cdot 2+3\cdot 3-4\cdot (-4) \\
& \Rightarrow 4+9+16 \\
& \Rightarrow 29 \\
\end{align}\]
As it is not equal to zero, hence, this is not the right combination.
Now, for (B), we get
\[\begin{align}
& \Rightarrow 2a+3b-4c \\
& \Rightarrow 2\cdot 4+3\cdot 4-4\cdot (5) \\
& \Rightarrow 8+12-20 \\
& \Rightarrow 0 \\
\end{align}\]
As it is equal to zero, hence, this is the right combination and hence, the correct answer is (B).
NOTE: An alternative method to solve this question is that we can take the cross product of the two vectors and as we know that the cross product of two perpendicular vectors is just the multiplication of the magnitudes of the two vectors and then, we can just put the values from the options and hence, find the correct answer.
\[\overrightarrow{v}\cdot \overrightarrow{u}=0\]
(Where \[\overrightarrow{v}\ and\ \overrightarrow{u}\] are two perpendicular vectors)
Another thing is that to calculate a dot product, we do the following
Let \[\overrightarrow{v}=a\widehat{i}+b\widehat{j}+c\widehat{k}\ and\ \overrightarrow{u}=l\widehat{i}+m\widehat{j}+n\widehat{k}\] be two vectors of which we have to calculate the dot product.
Now, to calculate their dot product, we can write the following \[\]
\[\begin{align}
& \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)\ \cdot \left( l\widehat{i}+m\widehat{j}+n\widehat{k} \right) \\
& \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=al+bm+cn \\
\end{align}\]
Complete step-by-step answer:
As mentioned in the question, vectors \[2\widehat{i}+3\widehat{j}-4\widehat{k}\ and\ a\widehat{i}+b\widehat{j}+c\widehat{k}\] are perpendicular to each other.
Now, as we know that the dot product of two vectors which are perpendicular is equal to 0, hence, we can find the required values of a, b and c as follows
\[\begin{align}
& \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)\ \cdot \left( l\widehat{i}+m\widehat{j}+n\widehat{k} \right)=al+bm+cn \\
& \Rightarrow \overrightarrow{v}\cdot \overrightarrow{u}=\left( 2\widehat{i}+3\widehat{j}-4\widehat{k} \right)\ \cdot \ \left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)=2a+3b-4c \\
\end{align}\]
Now, the dot product must be zero, so we can substitute the values of a, b and c from the options and then we can get the right combination.
Now, for (A), we get
\[\begin{align}
& \Rightarrow 2a+3b-4c \\
& \Rightarrow 2\cdot 2+3\cdot 3-4\cdot (-4) \\
& \Rightarrow 4+9+16 \\
& \Rightarrow 29 \\
\end{align}\]
As it is not equal to zero, hence, this is not the right combination.
Now, for (B), we get
\[\begin{align}
& \Rightarrow 2a+3b-4c \\
& \Rightarrow 2\cdot 4+3\cdot 4-4\cdot (5) \\
& \Rightarrow 8+12-20 \\
& \Rightarrow 0 \\
\end{align}\]
As it is equal to zero, hence, this is the right combination and hence, the correct answer is (B).
NOTE: An alternative method to solve this question is that we can take the cross product of the two vectors and as we know that the cross product of two perpendicular vectors is just the multiplication of the magnitudes of the two vectors and then, we can just put the values from the options and hence, find the correct answer.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Trending doubts
Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Giving reasons state the signs positive or negative class 12 physics CBSE

Explain esterification reaction with the help of a class 12 chemistry CBSE

What is defined as a solenoid Depict a diagram with class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Organisms of a higher trophic level which feed on several class 12 biology CBSE

