Name: Introduction to Vectors
Uploaded: 2019-07-19
Duration: 6 min 59 s
Description: Most of you might have seen vectors in Physics by now. Here, the same concept is explained through the lens of a Mathematician. Watch this video to learn more!!

Question

Show that the points with following position vectors are collinear:

$\hat{i} - 2 \hat{j} + 3 \hat{k}, 2 \hat{i} - 3 \hat{j} + 4 \hat{k}, - 2 \hat{i} + \hat{j}$

Accepted Answer

$3$ points $A, B, C$ are collinear if $|\vec{A B}| + |\vec{B C}| = |\vec{A C}|$

Let $\vec{O A} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{O B} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}, \vec{O C} = - 2 \hat{i} + \hat{j}$

$\vec{A B} = \vec{O B} - \vec{O A} =$ $2 \hat{i} - 3 \hat{j} + 4 \hat{k} - \hat{i} + 2 \hat{j} - 3 \hat{k} = \hat{i} - \hat{j} + \hat{k}$

$|\vec{A B}| = \sqrt{{(1)}^{2} + {(- 1)}^{2} + {(1)}^{2}} = \sqrt{3} u n i t s$

$\vec{B C} = \vec{O C} - \vec{O B} =$ $- 2 \hat{i} + \hat{j} - \hat{2 i} + 3 \hat{j} + 4 \hat{k} = - 4 \hat{i} + 4 \hat{j} + 4 \hat{k}$

$|\vec{B C}| = \sqrt{{(- 4)}^{2} + {(4)}^{2} + {(4)}^{2}} = 4 \sqrt{3} u n i t s$

$\vec{A C} = \vec{O C} - \vec{O A} =$ $- 2 \hat{i} + \hat{j} - \hat{i} + 2 \hat{j} - 3 \hat{k} = - 3 \hat{i} + 3 \hat{j} - 3 \hat{k}$

$|\vec{A C}| = \sqrt{{(- 3)}^{2} + {(3)}^{2} + {(- 3)}^{2}} = 3 \sqrt{3} u n i t s$

$\Rightarrow |\vec{A B}| + |\vec{B C}| = |\vec{A C}|$

Hence, the given points with position vectors

$\hat{i} - 2 \hat{j} + 3 \hat{k}, 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $- 2 \hat{i} + \hat{j}$ are collinear.

Show that the points with following position vectors are collinear:
$\hat{i} - 2 \hat{j} + 3 \hat{k}, 2 \hat{i} - 3 \hat{j} + 4 \hat{k}, - 2 \hat{i} + \hat{j}$

Important Questions on Vectors

Learn and Prepare for any exam you want !

Show that the points with following position vectors are collinear: i^-2j^+3k^, 2i^-3j^+4k^, -2i^+j^

Important Questions on Vectors

Learn and Prepare for any exam you want !

Show that the points with following position vectors are collinear:
$\hat{i} - 2 \hat{j} + 3 \hat{k}, 2 \hat{i} - 3 \hat{j} + 4 \hat{k}, - 2 \hat{i} + \hat{j}$