# If $\vec p = \hat i + a\hat j + \hat k$ and $\vec q = \hat i + \hat j + \hat k$, then $\left| {\vec p + \vec q} \right| = \left| {\vec p} \right| + \left| {\vec q} \right|$ is true for ?

A. $a = - 1$

B. $a = 1$

C. All real values of ‘a’

D. For no real values of ‘a’

Last updated date: 26th Mar 2023

•

Total views: 206.7k

•

Views today: 2.84k

Answer

Verified

206.7k+ views

**Hint:**In the given question, we are given the vector $\vec q$ in the form of rectangular directional coordinates. We are also given a vector $\vec p$ such that the component of the vector along the y axis is missing and is given to us as a constant a. now, we have to find the value of this missing constant such that the given condition $\left| {\vec p + \vec q} \right| = \left| {\vec p} \right| + \left| {\vec q} \right|$ holds true.

**Complete step by step answer:**

The given question revolves around the concepts of addition of two vectors and the magnitude of the resultant of two vectors. Now, we know that the magnitude of resultant of two vectors is $\sqrt {{{\left| {\vec a} \right|}^2} + {{\left| {\vec b} \right|}^2} + 2\left| {\vec a} \right|\left| {\vec b} \right|\cos \theta } $, where the angle $\theta $ is the angle between the two vectors $\vec a$ and $\vec b$.

So, finding the magnitude of the sum of the vectors $\vec p$ and $\vec q$, we get,

\[ \Rightarrow \left| {\vec p + \vec q} \right| = \sqrt {{{\left| {\vec p} \right|}^2} + {{\left| {\vec q} \right|}^2} + 2\left| {\vec p} \right|\left| {\vec q} \right|\cos \theta } \]

We are given that $\left| {\vec p + \vec q} \right| = \left| {\vec p} \right| + \left| {\vec q} \right|$.

So, we get,

\[ \Rightarrow \sqrt {{{\left| {\vec p} \right|}^2} + {{\left| {\vec q} \right|}^2} + 2\left| {\vec p} \right|\left| {\vec q} \right|\cos \theta } = \left| {\vec p} \right| + \left| {\vec q} \right|\]

We know that \[\left| {\vec p} \right| + \left| {\vec q} \right| = \left( {\sqrt {{{\left| {\vec p} \right|}^2} + {{\left| {\vec q} \right|}^2} + 2\left| {\vec p} \right|\left| {\vec q} \right|} } \right)\]. So, simplifying the right side of the equation, we get,

\[ \Rightarrow \sqrt {{{\left| {\vec p} \right|}^2} + {{\left| {\vec q} \right|}^2} + 2\left| {\vec p} \right|\left| {\vec q} \right|\cos \theta } = \sqrt {{{\left| {\vec p} \right|}^2} + {{\left| {\vec q} \right|}^2} + 2\left| {\vec p} \right|\left| {\vec q} \right|} \]

Squaring both sides and comparing both sides of the equation, we get,

\[ \Rightarrow {\left| {\vec p} \right|^2} + {\left| {\vec q} \right|^2} + 2\left| {\vec p} \right|\left| {\vec q} \right|\cos \theta = {\left| {\vec p} \right|^2} + {\left| {\vec q} \right|^2} + 2\left| {\vec p} \right|\left| {\vec q} \right|\]

Cancelling the like terms with opposite signs, we get,

\[ \Rightarrow 2\left| {\vec p} \right|\left| {\vec q} \right|\cos \theta = 2\left| {\vec p} \right|\left| {\vec q} \right|\]

Simplifying the equation, we get,

\[ \Rightarrow \cos \theta = 1\]

So, we get the value of $\cos \theta $ as $1$. So, this means that the two vectors $\vec p$ and $\vec q$ have an angle equal to zero degree between them.

This means that the two vectors are in the same direction. So, the direction cosines of two vectors should be equal. So, the directional cosines of the vector $\vec p$ are:

$\left( {\dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }},\dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }},\dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }}} \right)$

$ \Rightarrow \left( {\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}} \right)$

Now, the directional cosines of the vector $\vec p$ are: $\left( {\dfrac{1}{{\sqrt {{1^2} + {a^2} + {1^2}} }},\dfrac{a}{{\sqrt {{1^2} + {a^2} + {1^2}} }},\dfrac{1}{{\sqrt {{1^2} + {a^2} + {1^2}} }}} \right)$

$ \Rightarrow \left( {\dfrac{1}{{\sqrt {2 + {a^2}} }},\dfrac{a}{{\sqrt {2 + {a^2}} }},\dfrac{1}{{\sqrt {2 + {a^2}} }}} \right)$

Now, equating both the directional cosines, we get,

$\dfrac{1}{{\sqrt {2 + {a^2}} }} = \dfrac{1}{{\sqrt 3 }}$

$ \Rightarrow 2 + {a^2} = 3$

$ \Rightarrow {a^2} = 1$

$ \Rightarrow a = \pm 1$

Also, $\dfrac{a}{{\sqrt {2 + {a^2}} }} = \dfrac{1}{{\sqrt 3 }}$

This equation can only be true for $a = 1$. So, the value of a is $1$.

**Hence, option B is correct.**

**Note:**One must have a strong grip over the vectors and coordinate geometry in order to solve such questions. We should take care while handling the calculative steps in order to be sure of the answer. The given condition $\left| {\vec p + \vec q} \right| = \left| {\vec p} \right| + \left| {\vec q} \right|$ holds true only if the two vectors are ins= same direction. This result should be memorized so as to be used in other problems.

Recently Updated Pages

Which element possesses the biggest atomic radii A class 11 chemistry JEE_Main

The highly efficient method of obtaining beryllium class 11 chemistry JEE_Main

Which of the following sulphates has the highest solubility class 11 chemistry JEE_Main

Amongst the metal Be Mg Ca and Sr of group 2 of the class 11 chemistry JEE_Main

Which of the following metals is present in the greencolored class 11 chemistry JEE_Main

To prevent magnesium from oxidation in the electrolytic class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is the difference between anaerobic aerobic respiration class 10 biology CBSE