Ex 10.2 Class 12 Maths Ncert Solutions

\(
\overrightarrow{a}=\hat{\imath}+\hat{\jmath}+\hat{k}\)
\(\therefore|\overrightarrow{a}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}\)
\(\overrightarrow{b}=2 \hat{\imath}-7 \hat{\jmath}-3 \hat{k}\)
\(\therefore|\overrightarrow{b}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{2^{2}+(-7)^{2}+(-3)^{2}}=\sqrt{62}\)
\(\overrightarrow{c}=\frac{1}{\sqrt{3}} \hat{\imath}+\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k}\)
\(\therefore|\overrightarrow{c}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{-1}{\sqrt{3}}\right)^{2}}=\sqrt{1}=1
\)

Let \( \overrightarrow{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \)
And
\(
\overrightarrow{b}=\hat{\imath}-\hat{\jmath}-\hat{k}
\)
We can see clearly \( \overrightarrow{a} \neq \overrightarrow{b} \) because the all the coefficients of \( \overrightarrow{a} \) and \( \overrightarrow{b} \) are not same. In î and \( \hat{\jmath} \), the coefficients are different. Now, we check the magnitude of both,
\(
|\overrightarrow{a}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}
\)
And
\(
|\overrightarrow{a}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{1^{2}+(-1)^{2}+(-1)^{2}}=\sqrt{3}
\)
So, magnitude of both vectors is same but they are different.

Let \( \overrightarrow{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \)
And
\(
\overrightarrow{b}=2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k}=2(\hat{\imath}+\hat{\jmath}+\hat{k})=2 \overrightarrow{a}
\)
So, here \( \overrightarrow{b}=m \overrightarrow{a} \) where \( m=2 > 0 \)
Therefore, both vectors have same direction.
Now, we check for the magnitude, \( \overrightarrow{b}=2 \overrightarrow{a} \)
\(
|\overrightarrow{b}|=2|\overrightarrow{a}|
\)
So, they have same direction but same magnitude.
\(
\therefore \overrightarrow{a} \neq \overrightarrow{b}
\)

For two vectors to be equal, the coefficients of both vectors should be equal.
Comparing the \( \hat{\imath} \)-coefficient, we get \( x=2 \)
Comparing the \( \hat{\jmath} \)-coefficient, we get \( y=3 \)

Let A be the initial point \( (2,1) \) and B be the final point \( (-5,7) \).
So, \( \overrightarrow{A}=2 \hat{\imath}+\hat{\jmath} \) and \( \overrightarrow{B}=-5 \hat{\imath}+7 \hat{\jmath} \)
Now we want to find the \( \overrightarrow{A B} \)
\(
\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}=(-5 \hat{\imath}+7 \hat{\jmath})-(2 \hat{\imath}+\hat{\jmath})\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}=(-7 \hat{\imath}+6 \hat{\jmath})
\)
Therefore, the scalar components of \( \overrightarrow{\mathrm{AB}} \) are -7 and 6 . The vector components of \( \overrightarrow{\mathrm{AB}} \) are \( -7 \hat{\imath}+6 \hat{\jmath} \).

\(
\overrightarrow{a}=\hat{\imath}-2 \hat{\jmath}+\hat{k}\)
\(\overrightarrow{b}=-2 \hat{\imath}+4 \hat{\jmath}+5 \hat{k}
\)
And
\(
\vec{c}=\hat{\imath}-6 \hat{\jmath}-7 \hat{k}
\)
We want to find \( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} \)
To find the sum, we add coefficients of \( \hat{\imath}, \hat{\jmath}, \hat{k} \) to each other.
So, \( \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=(\hat{\imath}-2 \hat{\jmath}+\hat{k})+(-2 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})+(\hat{\imath}-6 \hat{\jmath}-7 \hat{k}) \)
\(
\Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0 \hat{\imath}-4 \hat{\jmath}-\hat{k}
\)

We know that unit vector means that the magnitude of the vector is 1 (unit).
It is defined as \( \hat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|} \)
So, we find the magnitude of \( \overrightarrow{a} \) first.
\(
\Rightarrow|\overrightarrow{a}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{6}\)
\(\hat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{(\hat{\imath}-2 \hat{\jmath}+\hat{k})}{\sqrt{6}}=\frac{1}{\sqrt{6}} \hat{\imath}+\frac{1}{\sqrt{6}} \hat{\jmath}+\frac{2}{\sqrt{6}} \hat{k}
\)

So, \( \overrightarrow{\mathrm{P}}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} \) and \( \overrightarrow{\mathrm{Q}}=4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k} \)
Now we want to find the \( \overrightarrow{P Q} \)
\(
\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{Q}}-\overrightarrow{\mathrm{P}}\)
\(\Rightarrow \overrightarrow{\mathrm{PQ}}=(4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})\)
\(\Rightarrow \overrightarrow{\mathrm{PQ}}=(3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k})
\)
Now, we have to find the unit direction in the direction of \( \overrightarrow{\mathrm{PQ}} \). We know that unit vector means that the magnitude of the vector is 1 (unit).
It is defined as \( \widehat{\mathrm{PQ}}=\frac{\overrightarrow{\mathrm{PQ}}}{|\mathrm{PQ}|} \)
So, we find the magnitude of \( \vec{a} \) first.
\(
\Rightarrow|\overrightarrow{\mathrm{PQ}}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{3^{2}+3^{2}+3^{2}}=\sqrt{27}=3 \sqrt{3}\)
\(\widehat{\mathrm{PQ}}=\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{QQ}}|}=\frac{(3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k})}{3 \sqrt{3}}=\frac{1}{\sqrt{3}} \hat{\imath}+\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k}
\)

\(
\overrightarrow{a}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k}\)
\(\overrightarrow{b}=-\hat{\imath}+\hat{\jmath}-\hat{k}
\)
We want to find \( \overrightarrow{a}+\overrightarrow{b} \)
To find the sum, we add coefficients of \( \hat{\imath}, \hat{\jmath}, \hat{k} \) to each other.
So,
\(
\overrightarrow{a}+\overrightarrow{b}=(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})+(-\hat{\imath}+\hat{\jmath}-\hat{k})\)
\(\Rightarrow \overrightarrow{a}+\overrightarrow{b}=1 \hat{\imath}-0 \hat{\jmath}+\hat{k}
\)
Now, we have to find the unit direction in the direction of \( (\overrightarrow{a}+\overrightarrow{b}) \). We know that unit vector means that the magnitude of the vector is 1 (unit).
So, we find the magnitude of \( \overrightarrow{a}+\overrightarrow{b} \) first.
\(
\Rightarrow|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{1^{2}+0^{2}+(1)^{2}}=\sqrt{2}\)
\(\frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}=\frac{(\hat{\imath}+0 \hat{\jmath}+\hat{k})}{\sqrt{2}}=\frac{1}{\sqrt{2}} \hat{\imath}+\frac{0}{\sqrt{2}} \hat{\jmath}+\frac{1}{\sqrt{2}} \hat{k}
\)

Let \( \overrightarrow{a}=5 \hat{\imath}-\hat{\jmath}+2 \hat{k} \)
The vector in the direction of \( \overrightarrow{a} \) having unit magnitude is \( \hat{a} \).
So, The vector in the direction of \( \overrightarrow{a} \) having magnitude 8 units \( =8 \hat{a} \).
\(
\Rightarrow|\vec{a}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{5^{2}+(-1)^{2}+2^{2}}=\sqrt{30}\)
\(8 \hat{a}=8 \frac{\overrightarrow{a}}{|\overrightarrow{a}|}=8 \frac{(5 \hat{\imath}-\hat{\jmath}+2 \hat{k})}{\sqrt{30}}=\frac{40}{\sqrt{30}} \hat{\imath}+\frac{8}{\sqrt{30}} \hat{\jmath}+\frac{16}{\sqrt{30}} \hat{k}
\)

We know that two vectors are collinear if they have the same direction or are parallel or anti-parallel. They can be expressed in the form \( \overrightarrow{b}= \) \( m \overrightarrow{a} \) where a and b are vectors and ' \( m \) ' is a scalar quantity.
From the question,
Let \( \overrightarrow{a}=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k} \) and \( \overrightarrow{b}=-4 \hat{\imath}+6 \hat{\jmath}-8 \hat{k} \)
Here, \( \overrightarrow{b}=-4 \hat{\imath}+6 \hat{\jmath}-8 \hat{k}=-2(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}) \)
\(
\Rightarrow \overrightarrow{b}=-2 \vec{a}
\)
So, \( \overrightarrow{b}=m \overrightarrow{a} \) where, \( m=-2 \).
\( \therefore \) The given two vectors are collinear.

The direction cosines of a vector are defined as the coefficients of \( \hat{1}, \hat{\jmath}, \hat{k} \) in the unit vector in the direction of the vector.
So, first we find the unit vector in the direction of the vector.
Let \( \overrightarrow{a}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} \)
\( \Rightarrow|\overrightarrow{a}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{14} \)
\( \hat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})}{\sqrt{14}}=\frac{1}{\sqrt{14}} \hat{\imath}+\frac{2}{\sqrt{14}} \hat{\jmath}+\frac{3}{\sqrt{14}} \hat{k} \)
Therefore. The direction cosines of the given vector are \( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \)

So, \( \overrightarrow{A}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \) and \( \overrightarrow{B}=-1 \hat{\imath}-2 \hat{\jmath}+1 \hat{k} \)
Now we want to find the \( \overrightarrow{A B} \)
\(
\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}=(-1 \hat{\imath}-2 \hat{\jmath}+1 \hat{k})-(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}=(-2 \hat{\imath}-4 \hat{\jmath}+4 \hat{k})
\)
Now, we have to find the direction cosines which are the coefficients of the unit vector in the direction of \( \overrightarrow{\mathrm{AB}} \). We know that unit vector means that the magnitude of the vector is 1 (unit).
It is defined as \( \widehat{\mathrm{AB}}=\frac{\overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AB}}|} \)
So, we find the magnitude of \( \overrightarrow{a} \) first.
\(
\Rightarrow|\overrightarrow{\mathrm{AB}}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{(-2)^{2}+(-4)^{2}+4^{2}}=\sqrt{36}=6\)
\(\widehat{\mathrm{AB}}=\frac{\overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AB}}|}=\frac{(-2 \hat{\imath}-4 \hat{\jmath}+4 \hat{k})}{6}=\frac{-1}{3} \hat{\imath}-\frac{2}{3} \hat{\jmath}+\frac{2}{3} \widehat{k}
\)
Therefore, The direction cosines of the given vector are \( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3} \).

To find the inclination of the vector with \( \mathrm{OX}, \mathrm{OY}, \mathrm{OZ} \). We find the direction cosines of the vector.
We know that the direction cosines of a vector are defined as the coefficients of \( \hat{1}, \hat{\jmath}, \hat{k} \) in the unit vector in the direction of the vector.
So, first we find the unit vector in the direction of the vector.
Let \( \overrightarrow{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \)
\(
\Rightarrow|\overrightarrow{a}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}\)
\(\hat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{(\hat{\imath}+\hat{\jmath}+\hat{k})}{\sqrt{3}}=\frac{1}{\sqrt{3}} \hat{\imath}+\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k}
\)
Therefore. The direction cosines of the given vector are \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \)
Let \( \theta_{1} \) be the angle between \( \overrightarrow{a} \) and OX.
Therefore, \( \cos \theta_{1}=\frac{1}{\sqrt{3}} \)
\(
\Rightarrow \theta_{1}=\cos ^{-1} \frac{1}{\sqrt{3}}
\)
Similarly, Let \( \theta_{2} \) be the angle between \( \overrightarrow{a} \) and OY.
Therefore, \( \cos \theta_{2}=\frac{1}{\sqrt{3}} \)
\(
\Rightarrow \theta_{2}=\cos ^{-1} \frac{1}{\sqrt{3}}
\)
And, \( \theta_{3} \) be the angle between \( \overrightarrow{a} \) and OZ.
Therefore, \( \cos \theta_{3}=\frac{1}{\sqrt{3}} \)
Therefore, \( \theta_{1}=\theta_{2}=\theta_{3} \)
Hence proved that the vector is equally inclined with the axes OX, OY, and OZ .

Given that \( \overrightarrow{\mathrm{P}}=\hat{\imath}+2 \hat{\jmath}-\hat{k} \) and \( \overrightarrow{\mathrm{Q}}=-\hat{\imath}+\hat{\jmath}+\hat{k} \)
(i) The \( \overrightarrow{R} \) lies on the segment PQ (internal division).If \( \mathrm{m}: \mathrm{n} \) is the ratio in which \( \overrightarrow{\mathrm{R}} \) divides PQ , then
\(
\overrightarrow{R}=\frac{m \overrightarrow{Q}+n \overrightarrow{P}}{m+n}
\)
Given \( \mathrm{m}: \mathrm{n}=2: 1, \mathrm{~m}=2 \) and \( \mathrm{n}=1 \)
\(
\Rightarrow \overrightarrow{\mathrm{R}}=\frac{2(-\hat{\imath}+\hat{\jmath}+\hat{k})+1(\hat{\imath}+2 \hat{\jmath}-\hat{k})}{2+1}=\frac{-1 \hat{\imath}+4 \hat{\jmath}+\hat{k}}{3}=-\frac{1}{3} \hat{\imath}+\frac{4}{3} \hat{\jmath}+\frac{1}{3} \hat{k}
\)
(ii) The \( \overrightarrow{R} \) does not lie on the segment PQ (external division).If \( \mathrm{m}: \mathrm{n} \) is the ratio in which divides PQ , then
\(
\overrightarrow{R}=\frac{\mathrm{m} \overrightarrow{\mathrm{Q}}-\mathrm{n} \overrightarrow{\mathrm{P}}}{\mathrm{m}-\mathrm{n}}
\)
Given \( \mathrm{m}: \mathrm{n}=2: 1, \mathrm{~m}=2 \) and \( \mathrm{n}=1 \)
\(
\Rightarrow \overrightarrow{\mathrm{R}}=\frac{2(-\hat{1}+\hat{\jmath}+\hat{k})-1(\hat{\mathrm{i}}+2 \hat{\jmath}-\hat{k})}{2-1}=\frac{-3 \hat{1}+0 \hat{\jmath}+3 \hat{k}}{1}=-3 \hat{\imath}+3 \hat{\mathrm{k}}
\)

So, \( \overrightarrow{\mathrm{P}}=2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k} \) and \( \overrightarrow{\mathrm{Q}}=4 \hat{\imath}+1 \hat{\jmath}-2 \hat{k} \)
Let \( \overrightarrow{\mathrm{R}} \) be the midpoint of \( \overrightarrow{\mathrm{PQ}} \).
\(
\overrightarrow{\mathrm{R}}=\frac{\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}}{2}\)
\(\Rightarrow \overrightarrow{\mathrm{R}}=\frac{(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})+(4 \hat{\imath}+1 \hat{\jmath}-2 \hat{k})}{2}\)
\(\Rightarrow \overrightarrow{\mathrm{R}}=\frac{(6 \hat{\imath}+4 \hat{\jmath}+2 \hat{k})}{2}\)
\(\Rightarrow \overrightarrow{\mathrm{R}}=3 \hat{\imath}+2 \hat{\jmath}+\hat{k}
\)

O be the origin,
Let \( \overrightarrow{a}=\overrightarrow{O A}=3 \hat{\imath}-4 \hat{\jmath}-4 \hat{k} \)
\(
\overrightarrow{b}=\overrightarrow{\mathrm{OB}}=2 \hat{\imath}-\hat{\jmath}+\hat{k}
\)
And
\(
\overrightarrow{c}=\overrightarrow{\mathrm{OC}}=\hat{\imath}-3 \hat{\jmath}-5 \hat{k}
\)
Now we find the vectors \( \overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CA}} \)
\(
\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\imath}-\hat{\jmath}+\hat{k})-(3 \hat{\imath}-4 \hat{\jmath}-4 \hat{k})\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}=(-\hat{\imath}+3 \hat{\jmath}+5 \hat{k})\ldots(1)\)
\(\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{C}}-\overrightarrow{\mathrm{B}}\)
\(\Rightarrow \overrightarrow{\mathrm{BC}}=(\hat{\imath}-3 \hat{\jmath}-5 \hat{k})-(2 \hat{\imath}-\hat{\jmath}+\hat{k})\)
\(\Rightarrow \overrightarrow{\mathrm{BC}}=(-\hat{\imath}-2 \hat{\jmath}-6 \hat{k})\ldots(2)\)
\(\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{C}}\)
\(\Rightarrow \overrightarrow{\mathrm{CA}}=(3 \hat{\imath}-4 \hat{\jmath}-4 \hat{k})-(\hat{\imath}-3 \hat{\jmath}-5 \hat{k})\)
\(\Rightarrow \overrightarrow{\mathrm{CA}}=(2 \hat{\imath}-\hat{\jmath}+\hat{k})\ldots(3)\)
Adding equations (1), (2) and (3)
\(
\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=(-\hat{\imath}+3 \hat{\jmath}+5 \hat{k})+(-\hat{\imath}-2 \hat{\jmath}-6 \hat{k})+(2 \hat{\imath}-\hat{\jmath}+\hat{k})\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k}
\)
Therefore, ABC form a triangle.
Now, we want to prove that it is a right angled triangle.
\(
|\overrightarrow{\mathrm{AB}}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{(-1)^{2}+(3)^{2}+5^{2}}=\sqrt{35}\)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{(-1)^{2}+(2)^{2}+(-6)^{2}}=\sqrt{41}\)
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{(2)^{2}+(-1)^{2}+1^{2}}=\sqrt{6}\)
\(\Rightarrow|\overrightarrow{\mathrm{AB}}|^{2}+|\overrightarrow{\mathrm{CA}}|^{2}=(\sqrt{35})^{2}+(\sqrt{6})^{2}=35+6=41\)
\(\Rightarrow|\overrightarrow{\mathrm{AB}}|^{2}+|\overrightarrow{\mathrm{CA}}|^{2}=41=(\sqrt{41})^{2}=|\overrightarrow{\mathrm{BC}}|^{2}
\)
Therefore, the triangle satisfies the Pythagoras theorem. Hence, proved the given vectors form a right angled triangle.

We know by triangle law of vectors,
\( \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}} \ldots (1)\)
Therefore, \( \Rightarrow \overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{A C}=0 \)
Hence \( (B) \) is true.
We know that \( \overrightarrow{\mathrm{AC}}=-\overrightarrow{\mathrm{CA}} \)
Putting in eq (1), we get
\(
\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=-\overrightarrow{\mathrm{CA}}\)
\(\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=0
\ldots (2)\)
Hence (A) is also true.
Now, \( \overrightarrow{\mathrm{BC}}=-\overrightarrow{\mathrm{CB}} \)
Putting in eq (2)
\(
\Rightarrow \overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=0
\)
Hence (D) is correct.
We are asked the option which is not true.
Therefore, the correct option is (C).

We know that two vectors are collinear if they have the same direction or are parallel or anti-parallel. They can be expressed in the form and \( \overrightarrow{b}= \) \( \mathrm{m} \overrightarrow{a} \) where a and b are vectors and ' m ' is a scalar quantity.
Therefore, (a) is true.
In (b), \( \mathrm{m}= \pm 1 \)
So, (b) is also true.
The vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \) are proportional,
Therefore, (c) is not true.
The vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \) can have different magnitude as well as different direction.
Therefore, (d) is not true.
We are asked the options which are not true.
\( \therefore \) The correct answer is (c) and (d).