# If a and b are unit vectors and $\left| a+b \right|=1,$then $\left| a-b \right|$ is equal to:

(a) $\sqrt{2}$

(b) $\sqrt{1}$

(c)$\sqrt{5}$

(d) $\sqrt{3}$

Answer

Verified

361.8k+ 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 $\left| a-b \right|=\sqrt{{{a}^{2}}+{{b}^{2}}-a.b}$

Last updated date: 24th Sep 2023

â€¢

Total views: 361.8k

â€¢

Views today: 8.61k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the basic unit of classification class 11 biology CBSE

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Draw a welllabelled diagram of a plant cell class 11 biology CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE