Question

Show that the points 0(0,0,0), A(2,-3, 3), B(-2,3,-3) are collinear. Find the ratio in which each point divides the segment joining the other two ​

Answer 1

Answer:

Given:

0(0,0,0), A(2, -3, 3), B(-2, 3, -3) are collinear

Let's find area of △OAB which is given by

\Delta=\frac{1}{2}|\mathrm{OA}||\mathrm{OB}| \sin \thetaΔ=

2

1

∣OA∣∣OB∣sinθ

\text { where } \theta \text { is angle between } \overrightarrow{\mathrm{OA}} \text { and } \overrightarrow{\mathrm{OB}} where θ is angle between

OA

and

OB

\text { Now, } \cos \theta=\frac{\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OA}}}{|\overrightarrow{\mathrm{OA}}||\overrightarrow{\mathrm{OB}}|} Now, cosθ=

∣

OA

∣∣

OB

∣

OA

⋅

OA

=\frac{-4-9-9}{\sqrt{4+9+9} \sqrt{4+9+9}}=\frac{-22}{22}=

4+9+9

4+9+9

−4−9−9

=

22

−22

\begin{gathered}\begin{array}{l}\cos \theta=-1 \Rightarrow \theta=180^{\circ} \\\Delta=0 \text { as } \sin 180=0\end{array}\end{gathered}

cosθ=−1⇒θ=180

∘

Δ=0 as sin180=0

O,A,B are collinear

\text { Now, let A divides } \overrightarrow{\mathrm{OB}} \text { in } \mathrm{k}: 1 Now, let A divides

OB

in k:1

2=\frac{-2 \mathrm{k}+0}{\mathrm{k}+1}2=

k+1

−2k+0

2 \mathrm{k}+2=-2 \mathrm{k}2k+2=−2k

4 \mathrm{k}=-2 \Rightarrow \mathrm{k}=-\frac{1}{2}4k=−2⇒k=−

2

1

\text { A divides } \overrightarrow{O B} \text { in } 1: 2 \text { externally } A divides

OB

in 1:2 externally

Now for B

-2=\frac{2 k+0}{k+1}−2=

k+1

2k+0

-2 \mathrm{k}-2=2 \mathrm{k}−2k−2=2k

4 \mathrm{k}=-24k=−2

\mathrm{k}=-\frac{1}{2}k=−

2

1

\text { B divides } \overrightarrow{O A} \text { in } 1: 2 \text { externally for O } B divides

OA

in 1:2 externally for O

\mathrm{O}=\frac{-2 \mathrm{k}+2}{\mathrm{k}+1}O=

k+1

−2k+2

2 \mathrm{k}=2 \Rightarrow \mathrm{k}=12k=2⇒k=1

\mathrm{O} \text { divides } \overrightarrow{\mathrm{AB}} \text { in } 1: 1 \text { internally }O divides

AB

in 1:1 internally

Hence the answer is \mathrm{O} \text { divides } \overrightarrow{\mathrm{AB}} \text { in } 1: 1 \text { internally }O divides

AB

in 1:1 internally

Answer 2

✔✏✏✏✏✏

Let's find area of △OAB which is given by,

Δ=

2

1

∣OA∣∣OB∣sinθ

where θ is angle between

OA

and

OB

Now, cosθ=

∣

∣

∣

∣

OA

∣

∣

∣

∣

∣

∣

∣

∣

OB

∣

∣

∣

∣

OA

.

OA

=

4+9+9

4+9+9

−4−9−9

=

22

−22

⇒ cosθ=−1⇒ θ=180

o

⇒ Δ=0 as sin180=0

⇒ O,A,B are collinear

Now, let A divides

OB

in k:1

⇒ 2=

k+1

−2k+0

⇒ 2k+2=−2k⇒ 4k=−2⇒ k=−

2

1

⇒ A divides

OB

in 1:2 externally

Now for B

⇒ −2=

k+1

2k+0

⇒ −2k−2=2k⇒ 4k=−2

⇒ k=−

2

1

⇒ B divides

OA

in 1:2 externally for O

⇒ O=

k+1

−2k+2

⇒ 2k=2⇒ k=1

⇒ O divides

AB

in 1:1 internally