Question

I: - Let $\overset{\to }{\mathop{\alpha }}\,=\left( x+4y \right)\overset{\to }{\mathop{a}}\,+\left( 2x+y+1 \right)\overset{\to }{\mathop{b}}\,,\overset{\to }{\mathop{\beta }}\,=\left( y-2x+2 \right)\overset{\to }{\mathop{a}}\,+\left( 2x-3y-1 \right)\overset{\to }{\mathop{b}}\,$, where a and b are non – zero, non – collinear vectors, if $3\overset{\to }{\mathop{\alpha }}\,=2\overset{\to }{\mathop{\beta }}\,$.
$\Rightarrow$ x = 2, y = -1.
II: - Let D, E, F be the middle points of the sides BC, CA, AB respectively of $\Delta ABC$ then, $\overset{\to }{\mathop{AD}}\,+\overset{\to }{\mathop{BE}}\,+\overset{\to }{\mathop{CF}}\,=0$.
(i) Only I is true.
(ii) Only II is true.
(iii) Both (i) and (ii) are true.
(iv) Neither (i) nor (ii) are true.

Answer 1

For statement I, follow the given condition: - $3\overset{\to }{\mathop{\alpha }}\,=2\overset{\to }{\mathop{\beta }}\,$ and equate then so that the coefficients of vectors a and b are equal. Now, form two linear equations and solve them to check if x = 2 and y = -1.
For statement II, draw a figure as per the given conditions and use triangle law of vector addition to get the result. Triangle law says that if head of ${{1}^{st}}$ vector is joined with the tail of ${{2}^{nd}}$ vector then the resultant vector is directed towards the head of ${{2}^{nd}}$ vector starting from the tail of ${{1}^{st}}$ vector.

Complete step by step answer:
(1) Let us consider statement I.
We have been given: - $\overset{\to }{\mathop{\alpha }}\,=\left( x+4y \right)\overset{\to }{\mathop{a}}\,+\left( 2x+y+1 \right)\overset{\to }{\mathop{b}}\,$ and $\overset{\to }{\mathop{\beta }}\,=\left( y-2x+2 \right)\overset{\to }{\mathop{a}}\,+\left( 2x-3y-1 \right)\overset{\to }{\mathop{b}}\,$. Here, we have been provided with the condition: - $3\overset{\to }{\mathop{\alpha }}\,=2\overset{\to }{\mathop{\beta }}\,$ and the values of x and y are to be matched with the given ones.
$\because$ $3\overset{\to }{\mathop{\alpha }}\,=2\overset{\to }{\mathop{\beta }}\,$

& \Rightarrow 3\left( x+4y \right)\overset{\to }{\mathop{a}}\,+3\left( 2x+y+1 \right)\overset{\to }{\mathop{b}}\,=2\left( y-2x+2 \right)\overset{\to }{\mathop{a}}\,+2\left( 2x-3y-1 \right)\overset{\to }{\mathop{b}}\, \\\ & \Rightarrow \left( 3x+12y \right)\overset{\to }{\mathop{a}}\,+\left( 6x+3y+3 \right)\overset{\to }{\mathop{b}}\,=\left( 2y-4x+4 \right)\overset{\to }{\mathop{a}}\,+\left( 4x-6y-2 \right)\overset{\to }{\mathop{b}}\, \\\ \end{aligned}$$ Now if the above equality holds true then the coefficients of both the vectors $$\overset{\to }{\mathop{a}}\,$$ and $$\overset{\to }{\mathop{b}}\,$$ must be equal. Therefore, equating the coefficients we get, $$\Rightarrow 3x+12y=2y-4x+4$$ and $$6x+3y+3=4x-6y-2$$. $$\Rightarrow 7x+10y=4$$ and $$2x+9y=-5$$ Considering $$7x+10y=4$$ as equation (i) and $$2x+9y=-5$$ as equation (ii), we get, $$7x+10y=4$$ - (i) $$2x+9y=-5$$ - (ii) Now, multiplying equation (i) with 2 and equation (ii) with 7 and then subtracting them to eliminate x, we get, $$\left( i \right)\Rightarrow 14x+20y=8$$ - (iii) $$\left( ii \right)\Rightarrow 14x+63y=-35$$ - (iv) Subtracting equation (iv) from equation (iii), we get, $$\begin{aligned} & \Rightarrow -43y=43 \\\ & \Rightarrow y=-1 \\\ \end{aligned}$$ Substituting y = -1, in equation (i), we get, $$\begin{aligned} & \Rightarrow 7x-10=4 \\\ & \Rightarrow 7x=14 \\\ & \Rightarrow x=2 \\\ \end{aligned}$$ Hence, it is clear that statement I is true. (2) Let us consider statement II. We have been given the D, E, F are the middle points of the sides BC, CA and AB respectively. Let us draw a diagram first so that we can visualize the question.  We have to find the value of expression, $$\overset{\to }{\mathop{AD}}\,+\overset{\to }{\mathop{BE}}\,+\overset{\to }{\mathop{CF}}\,=0$$. Using triangle law of vector addition which says that: - if head of $${{1}^{st}}$$ vector is joined with that tail of $${{2}^{nd}}$$ vector then the resultant vector is directed towards the head of $${{2}^{nd}}$$ vector starting from the tail of $${{1}^{st}}$$ vector. Considering, $$\Delta ABD$$,  $$\Rightarrow $$ $$\overset{\to }{\mathop{AD}}\,=\overset{\to }{\mathop{AB}}\,+\overset{\to }{\mathop{BD}}\,$$ Since, D is the mid – point of BC, therefore, $$\Rightarrow \overset{\to }{\mathop{AD}}\,=\overset{\to }{\mathop{AB}}\,+\dfrac{\overset{\to }{\mathop{BC}}\,}{2}$$ - (1) Considering, $$\Delta BEC$$,  $$\Rightarrow \overset{\to }{\mathop{BE}}\,=\overset{\to }{\mathop{BC}}\,+\overset{\to }{\mathop{CE}}\,$$ Since, E is the mid – point of AC, therefore, $$\Rightarrow \overset{\to }{\mathop{BE}}\,=\overset{\to }{\mathop{BC}}\,+\dfrac{\overset{\to }{\mathop{CA}}\,}{2}$$ - (ii) Considering, $$\Delta ACF$$,  $$\Rightarrow \overset{\to }{\mathop{CF}}\,=\overset{\to }{\mathop{CA}}\,+\overset{\to }{\mathop{AF}}\,$$ Since, F is the mid – point of AB, therefore, $$\Rightarrow \overset{\to }{\mathop{CF}}\,=\overset{\to }{\mathop{CA}}\,+\dfrac{\overset{\to }{\mathop{AB}}\,}{2}$$ - (iii) Adding equations (i), (ii) and (iii), we get, $$\Rightarrow \overset{\to }{\mathop{AB}}\,+\overset{\to }{\mathop{BE}}\,+\overset{\to }{\mathop{CF}}\,=\left( \overset{\to }{\mathop{AB}}\,+\overset{\to }{\mathop{BC}}\,+\overset{\to }{\mathop{CA}}\, \right)+\dfrac{1}{2}\left( \overset{\to }{\mathop{AB}}\,+\overset{\to }{\mathop{BC}}\,+\overset{\to }{\mathop{CA}}\, \right)$$ Now, in $$\Delta ABC$$,  $$\begin{aligned} & \overset{\to }{\mathop{\Rightarrow AB}}\,+\overset{\to }{\mathop{BC}}\,=\overset{\to }{\mathop{AC}}\,=-\overset{\to }{\mathop{CA}}\, \\\ & \Rightarrow \overset{\to }{\mathop{AB}}\,+\overset{\to }{\mathop{BC}}\,+\overset{\to }{\mathop{CA}}\,=0 \\\ \end{aligned}$$ Substituting this value in the above expression, we get, $$\begin{aligned} & \Rightarrow \overset{\to }{\mathop{AD}}\,+\overset{\to }{\mathop{BE}}\,+\overset{\to }{\mathop{CF}}\,=0+\dfrac{1}{2}\left( 0 \right) \\\ & \Rightarrow \overset{\to }{\mathop{AD}}\,+\overset{\to }{\mathop{BE}}\,+\overset{\to }{\mathop{CF}}\,=0 \\\ \end{aligned}$$ Hence, statement II is also true. **So, the correct answer is “Option (iii)”.** **Note:** One may note that, in standard II, we have been provided with $$\overset{\to }{\mathop{AD}}\,+\overset{\to }{\mathop{BE}}\,+\overset{\to }{\mathop{CF}}\,$$. These are vectors and not scalar. So we have to apply vector addition property and not simple scalar addition. We can also use parallelogram law of vector addition instead of triangle law but then the construction of triangle will become somewhat confusing and that’s why triangle law is used.