If $a.\hat i = a.(\hat j + \hat i) = a.(\hat i + \hat j + \hat k)$ , then the value of a is equal to?
Answer
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Hint: In this question, we have to find the vector $\overrightarrow a $ . To proceed Let vector $\overrightarrow a $ be $x\hat i + y\hat j + z\hat k$ and put it in the given equation to get the answer using the properties of the dot product.
Complete step-by-step answer:
Let $\overrightarrow a = x\hat i + y\hat j + z\hat k$
\[ \Rightarrow a.\hat i = (x\hat i + y\hat j + z\hat k).\hat i = x\]
It’s because we know that in dot product $\hat i.\hat i = 1$ as in dot product we have \[\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta \] where $\theta $ is the angle between vectors $\overrightarrow a $ and $\overrightarrow b $ , here value of $\theta $ is ${0^0}$ and we know that $\cos {0^0} = 1$ and also they are unit vectors so their magnitude is 1.
We also know that in dot product $\hat i.\hat j = 0$ as in dot product we have \[\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta \] where $\theta $ is the angle between vectors $\overrightarrow a $ and $\overrightarrow b $ , here value of $\theta $ is ${90^0}$ and we know that $\cos {90^0} = 0$ and also they are unit vectors so their magnitude is 1.
We also know that in dot product $\hat i.\hat k = 0$ as in dot product we have \[\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta \] where $\theta $ is the angle between vectors $\overrightarrow a $ and $\overrightarrow b $ , here value of $\theta $ is ${90^0}$ and we know that $\cos {90^0} = 0$ and also they are unit vectors so their magnitude is 1.
\[ \Rightarrow a.(\hat i + \hat j) = (x\hat i + y\hat j + z\hat k).(\hat i + \hat j) = x + y\]
And also,
\[ \Rightarrow a.(\hat i + \hat j + \hat k) = (x\hat i + y\hat j + z\hat k).(\hat i + \hat j + \hat k) = x + y + z\]
According to the question,
$ \Rightarrow a.\hat i = a.(\hat j + \hat i) = a.(\hat i + \hat j + \hat k)$
On substituting the values of each from the above results, we get
$ \Rightarrow x = x + y = x + y + z$
On solving the first two we get,
$ \Rightarrow x = x + y$
$ \Rightarrow y = 0$
Solving the second and third we get,
$ \Rightarrow x + y = x + y + z$
$ \Rightarrow z = 0$
From above we get x = 1
So, on substituting these in vector a we get
$ \Rightarrow a = \hat i$
Hence, we can say that the value of $a = \hat i$
Note- For these types of questions, we have to suppose a vector a in the form of variables at the start to proceed further. Also keep in mind the properties of cross and dot product of vectors for such questions.
Complete step-by-step answer:
Let $\overrightarrow a = x\hat i + y\hat j + z\hat k$
\[ \Rightarrow a.\hat i = (x\hat i + y\hat j + z\hat k).\hat i = x\]
It’s because we know that in dot product $\hat i.\hat i = 1$ as in dot product we have \[\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta \] where $\theta $ is the angle between vectors $\overrightarrow a $ and $\overrightarrow b $ , here value of $\theta $ is ${0^0}$ and we know that $\cos {0^0} = 1$ and also they are unit vectors so their magnitude is 1.
We also know that in dot product $\hat i.\hat j = 0$ as in dot product we have \[\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta \] where $\theta $ is the angle between vectors $\overrightarrow a $ and $\overrightarrow b $ , here value of $\theta $ is ${90^0}$ and we know that $\cos {90^0} = 0$ and also they are unit vectors so their magnitude is 1.
We also know that in dot product $\hat i.\hat k = 0$ as in dot product we have \[\overrightarrow a .\overrightarrow b = \left| a \right|\left| b \right|\cos \theta \] where $\theta $ is the angle between vectors $\overrightarrow a $ and $\overrightarrow b $ , here value of $\theta $ is ${90^0}$ and we know that $\cos {90^0} = 0$ and also they are unit vectors so their magnitude is 1.
\[ \Rightarrow a.(\hat i + \hat j) = (x\hat i + y\hat j + z\hat k).(\hat i + \hat j) = x + y\]
And also,
\[ \Rightarrow a.(\hat i + \hat j + \hat k) = (x\hat i + y\hat j + z\hat k).(\hat i + \hat j + \hat k) = x + y + z\]
According to the question,
$ \Rightarrow a.\hat i = a.(\hat j + \hat i) = a.(\hat i + \hat j + \hat k)$
On substituting the values of each from the above results, we get
$ \Rightarrow x = x + y = x + y + z$
On solving the first two we get,
$ \Rightarrow x = x + y$
$ \Rightarrow y = 0$
Solving the second and third we get,
$ \Rightarrow x + y = x + y + z$
$ \Rightarrow z = 0$
From above we get x = 1
So, on substituting these in vector a we get
$ \Rightarrow a = \hat i$
Hence, we can say that the value of $a = \hat i$
Note- For these types of questions, we have to suppose a vector a in the form of variables at the start to proceed further. Also keep in mind the properties of cross and dot product of vectors for such questions.
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