Question

If $a,b$ and $c$ are different real numbers and $a\hat{i}+b\hat{j}+c\hat{k},b\hat{i}+c\hat{j}+a\hat{k}$ and $c\hat{i}+a\hat{j}+b\hat{k}$ are position vectors of three non-collinear points, then?

• A. centroid of $\Delta ABC$ is $\frac{a+b+c}{3}(\hat{i}+\hat{j}+\hat{k})$
• B. $(\hat{i}+\hat{j}+\hat{k})$ is not equally inclined to three vectors
• C. $∆ABC$ is a scalene triangle
• D. perpendicular from the origin to the plane of the triangle does not meet it at the centroid

Answer 1

The correct option is A

centroid of $\Delta ABC$ is $\frac{a+b+c}{3}(\hat{i}+\hat{j}+\hat{k})$

Explanation of the correct option:

Step 1. Given that $a\hat{i}+b\hat{j}+c\hat{k},b\hat{i}+c\hat{j}+a\hat{k}$ and $c\hat{i}+a\hat{j}+b\hat{k}$ are position vectors of three non-collinear points

Now, As we know,

$\mathit{G}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\left(}\frac{\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}\mathbf{+}\overrightarrow{\mathbf{c}}}{\mathbf{3}}\mathbf{\right)}$

Step 2. The centroid of $\begin{array}{rcl}\mathbf{△}\mathbf{}\mathit{A}\mathit{B}\mathit{C}& =& \frac{\left(a\hat{i}+b\hat{j}+c\hat{k}\right)+\left(b\hat{i}+c\hat{j}+\hspace{0.17em}a\hat{k}\right)+\left(c\hat{i}+a\hat{j}+b\hat{k}\right)}{3}\end{array}$

$\begin{array}{rcl}& =& \frac{\left(a+b+c\right)\hat{i}+\left(a+b+c\right)\hat{j}+\left(a+b+c\right)\hat{k}}{3}\\ & =& \frac{\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}\mathbf{b}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}}{\mathbf{3}}\mathbf{(}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{}\hat{\mathbf{j}}\mathbf{}\mathbf{+}\mathbf{}\hat{\mathbf{k}}\mathbf{)}\end{array}$

Therefore, option (A) is correct.

Explanation for the incorrect option:

Step 3. Check if $(\hat{i}+\hat{j}+\hat{k})$ is not equally inclined to three vectors:

Option (B)

Let the given position vectors be of points $A,B$ and $C$ respectively, then

$\left|A\right|=\sqrt{a^2+b^2+c^2},\left|B\right|=\sqrt{a^2+b^2+c^2},\left|C\right|=\sqrt{a^2+b^2+c^2}$

Therefore, the direction cosines of $A,B$ and $C$ are

If $A$ makes angle $\alpha$ with $(\hat{i}+\hat{j}+\hat{k})$ then, we have

$\begin{array}{rcl}\cos \alpha & =& \frac{\left(ai+bj+ck\right)·\left(\hat{i}+\hat{j}+\hat{k}\right)}{\left|ai+bj+ck\right|\left|\hat{i}+\hat{j}+\hat{k}\right|}\\ & =& \frac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\end{array}$

$[\because$cosine of the angle between any two vectors is their dot product divided by the product of their magnitudes$]$

If $B$ makes angle $\beta$ with $(\hat{i}+\hat{j}+\hat{k})$ then, we have

$\begin{array}{rcl}\cos \beta & =& \frac{\left(bi+cj+ak\right)·\left(\hat{i}+\hat{j}+\hat{k}\right)}{\left|bi+cj+ak\right|\left|\hat{i}+\hat{j}+\hat{k}\right|}\\ & =& \frac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\end{array}$

If $C$ makes angle $\gamma$ with $(\hat{i}+\hat{j}+\hat{k})$ then, we have

$\begin{array}{rcl}\cos \gamma & =& \frac{\left(ci+aj+bk\right)·\left(\hat{i}+\hat{j}+\overrightarrow{k}\right)}{\left|ci+aj+bk\right|\left|\hat{i}+\hat{j}+\hat{k}\right|}\\ & =& \frac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\end{array}$

Now, $\alpha =\beta =\gamma ={\cos }^{-1}\left[\frac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\right]$

So, $(\hat{i}+\hat{j}+\hat{k})$ is equally inclined to three vectors

Step 4. Check if $∆ABC$ is a scalene triangle

Option (C)

$\left|\overrightarrow{AB}\right|=\sqrt{{\left(b-a\right)}^2+{\left(c-b\right)}^2+{\left(a-c\right)}^2}$

$\left|\overrightarrow{BC}\right|=\sqrt{{\left(c-b\right)}^2+{\left(a-c\right)}^2+{\left(a-b\right)}^2}$

$\left|\overrightarrow{CA}\right|=\sqrt{{\left(a-c\right)}^2+{\left(b-a\right)}^2+{\left(c-b\right)}^2}$

$\therefore \left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|=\left|\overrightarrow{CA}\right|$

So, $△ABC$ is an equilateral triangle

Option (D):

G vector is not perpendicular to the plane of triangle.

Hence, Option ‘A’ is Correct.