Answer
Verified
422.1k+ views
Hint: We can use the formula for modulus of vectors as $\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$ and $\left| a-b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}-a.b}$ to solve this question. Also the modulus of a unit vector is 1.
Complete step-by-step answer:
Before proceeding to the solution we should know that the magnitude is also called the modulus or the length of the vector. Magnitude is represented by the length of the directed line segment. A unit vector is a vector of length 1. To obtain a unit vector in the direction of any vector we divide by its modulus. A unit vector is a vector of unit length, sometimes also called a direction vector. The unit vector is defined by \[\widehat{v}\].
\[\widehat{v}=\dfrac{v}{\left| v \right|}\],
Given, $\left| a+b \right|=1$
Using the formula $\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$ we get:
Since, a and b are unit vectors therefore, $\left| a \right|=1,\left| b \right|=1$ and $\cos \theta $ is the angle between the vectors a and b.
So, we can substitute the values in the above equation and we will get,
$\therefore 1=\sqrt{{{1}^{2}}+{{1}^{2}}+2\times 1\times 1\times \cos \theta }$
$\Rightarrow 1=\sqrt{2+2\cos \theta }$
Squaring both sides we get,
$\Rightarrow 2+2\cos \theta =1$
$\Rightarrow 2\cos \theta =-1$
$\Rightarrow \cos \theta =-\dfrac{1}{2}......(i)$
Now, we have to find out the value of $\left| a-b \right|$ therefore, we can use the formula for $\left| a-b \right|$.
Since, we know the values, we can substitute them in the equation and we will get,
$\therefore \left| a-b \right|=\sqrt{1+1-2\times 1\times 1\times \cos \theta }$
$\Rightarrow \left| a-b \right|=\sqrt{2-2\cos \theta }$
Then, we can substitute the value of$\cos \theta $ from equation (i), and we will get,
$\Rightarrow \left| a-b \right|=\sqrt{2-2\times -\dfrac{1}{2}}$
$\therefore \left| a-b \right|=\sqrt{3}$
Hence, the answer is option (d).
Note: It is important to use the cos theta term in the modulus equation. The dot product of two vectors along with the angle between them by cos theta should be used. If not, it would not be easy to solve the question. Be very carefully about the value of modulus, modulus can never be negative. Always remember the formula of modulus like:
$\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$, and $\left| a-b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}-a.b}$
Complete step-by-step answer:
Before proceeding to the solution we should know that the magnitude is also called the modulus or the length of the vector. Magnitude is represented by the length of the directed line segment. A unit vector is a vector of length 1. To obtain a unit vector in the direction of any vector we divide by its modulus. A unit vector is a vector of unit length, sometimes also called a direction vector. The unit vector is defined by \[\widehat{v}\].
\[\widehat{v}=\dfrac{v}{\left| v \right|}\],
Given, $\left| a+b \right|=1$
Using the formula $\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$ we get:
Since, a and b are unit vectors therefore, $\left| a \right|=1,\left| b \right|=1$ and $\cos \theta $ is the angle between the vectors a and b.
So, we can substitute the values in the above equation and we will get,
$\therefore 1=\sqrt{{{1}^{2}}+{{1}^{2}}+2\times 1\times 1\times \cos \theta }$
$\Rightarrow 1=\sqrt{2+2\cos \theta }$
Squaring both sides we get,
$\Rightarrow 2+2\cos \theta =1$
$\Rightarrow 2\cos \theta =-1$
$\Rightarrow \cos \theta =-\dfrac{1}{2}......(i)$
Now, we have to find out the value of $\left| a-b \right|$ therefore, we can use the formula for $\left| a-b \right|$.
Since, we know the values, we can substitute them in the equation and we will get,
$\therefore \left| a-b \right|=\sqrt{1+1-2\times 1\times 1\times \cos \theta }$
$\Rightarrow \left| a-b \right|=\sqrt{2-2\cos \theta }$
Then, we can substitute the value of$\cos \theta $ from equation (i), and we will get,
$\Rightarrow \left| a-b \right|=\sqrt{2-2\times -\dfrac{1}{2}}$
$\therefore \left| a-b \right|=\sqrt{3}$
Hence, the answer is option (d).
Note: It is important to use the cos theta term in the modulus equation. The dot product of two vectors along with the angle between them by cos theta should be used. If not, it would not be easy to solve the question. Be very carefully about the value of modulus, modulus can never be negative. Always remember the formula of modulus like:
$\left| a+b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+a.b}$, and $\left| a-b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}-a.b}$
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Which of the following does not have the dimensions class 11 physics CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
What organs are located on the left side of your body class 11 biology CBSE