Exercise 8.1-Additional Questions - Chapter 8 - Vector Algebra – I - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Problems
Question 1.
Shown that the points with position vectors $\vec{a}-2 \vec{b}+3 \vec{c},-2 \vec{a}+3 \vec{b}+2 \vec{c}$ and $-8 \vec{a}+13 \vec{b}$ collinear.
Solution:
To prove the points $P, Q, R$ are collinear we have to prove that

$\overrightarrow{\mathrm{PQ}}=\mathrm{t} \overrightarrow{\mathrm{PR}}$ where $\mathrm{t}$ is a scalar.
Let the given points be $\mathrm{P}, \mathrm{Q}, \mathrm{R}$.
Then
$
\begin{aligned}
& \overrightarrow{\mathrm{OP}}=\vec{a}-2 \vec{b}+3 \vec{c} \\
& \overrightarrow{\mathrm{OQ}}=-2 \vec{a}+3 \vec{b}+2 \vec{c}
\end{aligned}
$
and
$
\overrightarrow{\mathrm{OR}}=-8 \vec{a}+13 \vec{b}
$
Now $\begin{aligned} \overrightarrow{\mathrm{PQ}} & =\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}=(-2 \vec{a}+3 \vec{b}+2 \vec{c})-(\vec{a}-2 \vec{b}+3 \vec{c}) \\ & =-2 \vec{a}+3 \vec{b}+2 \vec{c}-\vec{a}+2 \vec{b}-3 \vec{c} \\ & =-3 \vec{a}+5 \vec{b}-\vec{c} \\ \overrightarrow{\mathrm{PR}}=\overrightarrow{\mathrm{OR}}-\overrightarrow{\mathrm{OP}} & =(-8 \vec{a}+13 \vec{b})-(\vec{a}-2 \vec{b}+3 \vec{c}) \\ & =-8 \vec{a}+13 \vec{b}-\vec{a}+2 \vec{b}-3 \vec{c} \\ & =-9 \vec{a}+15 \vec{b}-3 \vec{c} \\ & =3(-3 \vec{a}+5 \vec{b}-\vec{c})=3 \overrightarrow{\mathrm{PQ}} \\ \overrightarrow{\mathrm{PR}} & =3 \overrightarrow{\mathrm{PQ}}\end{aligned}$
So, the points P, Q, R are collinear (i.e,) the given points are collinear.

Question 2 .
If $A B C$ and $A$ ' $B$ ' $C$ ' are two triangles and $\mathrm{G}, \mathrm{G}$ ' be their corresponding centroids, prove that
Solution:
Let $\mathrm{O}$ be the origin.
We know when $\mathrm{G}$ is the centroid of $\triangle \mathrm{ABC}$,
$
\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}=3 \overrightarrow{\mathrm{OG}}
$
Similarly when $\mathrm{G}^{\prime}$ is the centroid of $\Delta \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime}$,

Question 3.
Prove using vectors the mid-points of two opposite sides of a quadrilateral and the mid-points of the diagonals are the vertices of a parallelogram.
Solution:
$A B C D$ is a quadrilateral with position vectors
$\mathrm{OA}=\vec{a}, \mathrm{OB}=\vec{b}, \mathrm{OC}=\vec{c}$ and $\mathrm{OD}=\vec{d}$
$\mathrm{P}$ is the midpoint of $\mathrm{BC}$ and $\mathrm{R}$ is the midpoint of $\mathrm{AD}$.
$Q$ is the midpoint of $A C$ and $S$ is the midpoint of $B D$.

To prove $\mathrm{PQRS}$ is a parallelogram. We have to prove that $\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{SR}}$
Now

$
\begin{aligned}
& \overrightarrow{\mathrm{OP}}=\frac{\vec{b}+\vec{c}}{2} \\
& \overrightarrow{\mathrm{OR}}=\frac{\vec{a}+\vec{d}}{2} \\
& \overrightarrow{\mathrm{OQ}}=\frac{\vec{a}+\vec{c}}{2} \\
& \overrightarrow{\mathrm{OS}}=\frac{\vec{b}+\vec{d}}{2} \\
& \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}=\frac{\vec{a}+\vec{c}}{2}-\frac{\vec{b}+\vec{c}}{2}=\frac{\vec{a}-\vec{b}}{2} \\
& \overrightarrow{\mathrm{SR}}=\overrightarrow{\mathrm{OR}}-\overrightarrow{\mathrm{OS}}=\frac{\vec{a}+\vec{d}}{2}-\frac{\vec{b}+\vec{d}}{2}=\frac{\vec{a}-\vec{b}}{2} \\
& \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{SR}} \Rightarrow \mathrm{PQRS} \text { is a parallelogram. }
\end{aligned}
$