Question

If ABC is a triangle, then the vectors $\left( -1,\cos C,\cos B \right),\left( \cos C,-1,\cos A \right)\text{ and }\left( \cos B,\cos C,-1 \right)$ are:
(a) linearly independent for all triangles
(b) linearly dependent for all triangles
(c) linearly independent for all isosceles triangles
(d) none of these

Answer 1

In this question, we have given a triangle ABC with three vectors as $\left( -1,\cos C,\cos B \right),\left( \cos C,-1,\cos A \right)\text{ and }\left( \cos B,\cos C,-1 \right)$ and we have to find whether these three vectors are linearly dependent or independent. For that, we are going to find the determinant of the three vectors by putting the first vector in the first row of the determinant, second vector in the second row of the determinant and third vector in the third row of the determinant. After that we will find the determinant value of these three vectors. If the determinant value is 0 then the vectors are linearly dependent and if not 0 then the three vectors are linearly independent.

Complete step-by-step answer:
We have given a triangle ABC then the three vectors are given as:$\left( -1,\cos C,\cos B \right),\left( \cos C,-1,\cos A \right)\text{ and }\left( \cos B,\cos C,-1 \right)$
And we have to find whether these three vectors are linearly dependent or independent.
We know that, if the three vectors say:
$\begin{aligned} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z=A \\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z=B \\\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z=C \\\ \end{aligned}$
Then the determinant of the three vectors is equal to:
$\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|$
If the determinant is 0 then the three vectors are linearly dependent and if the determinant is non zero then the three vectors are linearly independent.
Now, the three vectors given in the above question as:

& \left( -1,\cos C,\cos B \right), \\\ & \left( \cos C,-1,\cos A \right)\text{,} \\\ & \left( \cos B,\cos C,-1 \right) \\\ \end{aligned}$$ Let us take one of the vectors $$\left( -1,\cos C,\cos B \right)$$ then in this vector: -1 is the projection of vector in x direction, $\cos C$ is the projection of vector in y direction and $\cos B$ is the projection of vector in z direction so the three vectors are written in the determinant form as: $\left| \begin{matrix} -1 & \cos C & \cos B \\\ \cos C & -1 & \cos A \\\ \cos B & \cos A & -1 \\\ \end{matrix} \right|$ Expanding the above determinant along the first row we get, $-1\left( \left| \begin{matrix} -1 & \cos A \\\ \cos A & -1 \\\ \end{matrix} \right| \right)-\cos C\left( \left| \begin{matrix} \cos C & \cos A \\\ \cos B & -1 \\\ \end{matrix} \right| \right)+\cos B\left( \left| \begin{matrix} \cos C & -1 \\\ \cos B & \cos A \\\ \end{matrix} \right| \right)$ Solving the above expression we get, $$\begin{aligned} & -1\left( -1\left( -1 \right)-\cos A\left( \cos A \right) \right)-\cos C\left( \cos C\left( -1 \right)-\cos A\cos B \right)+\cos B\left( \cos C\cos A-\left( -1 \right)\left( \cos B \right) \right) \\\ & =-1\left( 1-{{\cos }^{2}}A \right)-\cos C\left( -\cos C-\cos A\cos B \right)+\cos B\left( \cos C\cos A+\cos B \right) \\\ & =-1+{{\cos }^{2}}A+{{\cos }^{2}}C+\cos A\cos B\cos C+\cos A\cos B\cos C+{{\cos }^{2}}B \\\ & =-1+{{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C+2\cos A\cos B\cos C \\\ \end{aligned}$$ We know from the trigonometric identities that: ${{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1-2\cos A\cos B\cos C$ Using the above relation in $$-1+{{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C+2\cos A\cos B\cos C$$ we get, $$\begin{aligned} & -1+1-2\cos A\cos B\cos C+2\cos A\cos B\cos C \\\ & =0+0 \\\ & =0 \\\ \end{aligned}$$ From the above, we are getting the value of the determinant as 0. As the value of determinant of the three vectors is 0 so the three vectors are linearly dependent for all triangles. **So, the correct answer is “Option (b)”.** **Note:** In the above solution, we have written the trigonometric identity as follows: ${{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1-2\cos A\cos B\cos C$ In the below, we are going to show the derivation of the above identity, Solving the L.H.S of the above equation we get, ${{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C$ We know that, $2{{\cos }^{2}}A=1+\cos 2A$ Similarly, we can write ${{\cos }^{2}}B\And {{\cos }^{2}}C$. Multiplying the L.H.S by 2 we get, $2{{\cos }^{2}}A+2{{\cos }^{2}}B+2{{\cos }^{2}}C$ Substituting the value of $2{{\cos }^{2}}A,2{{\cos }^{2}}B$ that we have derived above in the above expression we get, $\begin{aligned} & 1+\cos 2A+1+\cos 2B+2{{\cos }^{2}}C \\\ & =2+\cos 2A+\cos 2B+2{{\cos }^{2}}C \\\ \end{aligned}$ We are going to apply the trigonometric identity of $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ in $\cos 2A+\cos 2B$ we get, $\begin{aligned} & 2+2\cos \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A+2B}{2} \right)+2{{\cos }^{2}}C \\\ & =2+2\cos \left( A+B \right)\cos \left( A-B \right)+2{{\cos }^{2}}C \\\ \end{aligned}$ We know that, $A+B+C=\pi $ Subtracting C on both the sides we get, $A+B=\pi -C$ Substituting $A+B=\pi -C$ in $2+2\cos \left( A+B \right)\cos \left( A-B \right)+2{{\cos }^{2}}C$ we get, $2+2\cos \left( \pi -C \right)\cos \left( A-B \right)+2{{\cos }^{2}}C$ We know that $\cos \left( \pi -C \right)=-\cos C$ so using this relation in the above expression we get, $\begin{aligned} & 2+2\left( -\cos C \right)\cos \left( A-B \right)+2{{\cos }^{2}}C \\\ & =2-2\cos C\left( -\cos C+\cos \left( A-B \right) \right) \\\ \end{aligned}$ Substituting $C=\pi -\left( A+B \right)$ in the above expression we get, $\begin{aligned} & 2-2\cos C\left( -\cos \left( \pi -\left( A+B \right) \right)+\cos \left( A-B \right) \right) \\\ & =2-2\cos C\left( +\cos \left( A+B \right)+\cos \left( A-B \right) \right) \\\ & =2-2\cos C\left( \cos \left( A+B \right)+\cos \left( A-B \right) \right) \\\ & =2-2\cos C\left( 2\cos \left( \dfrac{2A}{2} \right)\cos \left( \dfrac{2B}{2} \right) \right) \\\ & =2-4\cos A\cos B\cos C \\\ \end{aligned}$ Now, this is the expansion of $2{{\cos }^{2}}A+2{{\cos }^{2}}B+2{{\cos }^{2}}C$ so to get the value for ${{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C$ we have to divide the above expression by 2 we get, $1-2\cos A\cos B\cos C$ The above expression is equal to the R.H.S. Hence, we have proved.