Consider the position vectors of A, B and C as O⃗A⃗ = 2î - 2ĵ + k̂, O⃗B⃗ = î + 2ĵ - 2k̂ and O⃗C⃗ = 2î - ĵ + 4k̂ Calculate A⃗B⃗ and B⃗C⃗. Find the…

A⃗B⃗=O⃗B⃗-O⃗A⃗=-î+4ĵ-3k̂ B⃗C⃗=O⃗C⃗-O⃗B⃗=î-3ĵ+6k̂ A⃗B⃗=-î+4ĵ-3k̂, B⃗C⃗=î-3ĵ+6k̂

Mathematics20254 marksNumericalVectorsTool: 3D geometry and vectors calculator

Solve the following.

Consider the position vectors of A, B and C as O⃗A⃗ = 2î - 2ĵ + k̂, O⃗B⃗ = î + 2ĵ - 2k̂ and…

Consider the position vectors of $A$, $B$ and $C$ as $\vec{OA} = 2\hat{i} - 2\hat{j} + \hat{k}$, $\vec{OB} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{OC} = 2\hat{i} - \hat{j} + 4\hat{k}$

(i)[1]

Calculate $\vec{AB}$ and $\vec{BC}$.

(ii)[1]

Find the projection of $\vec{AB}$ on $\vec{BC}$.

(iii)[2]

Find the area of the triangle $ABC$ whose sides are $\vec{AB}$ and $\vec{BC}$.

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Answer

Answer (i)

$\vec{AB}=\vec{OB}-\vec{OA}=-\hat i+4\hat j-3\hat k$ $\vec{BC}=\vec{OC}-\vec{OB}=\hat i-3\hat j+6\hat k$

Final answer: $\vec{AB}=-\hat i+4\hat j-3\hat k,\ \vec{BC}=\hat i-3\hat j+6\hat k$

Answer (ii)

Projection $=\frac{\vec{AB}\cdot\vec{BC}}{|\vec{BC}|}=\frac{-1-12-18}{\sqrt{1+9+36}}=\frac{-31}{\sqrt{46}}$.

Final answer: $-\frac{31}{\sqrt{46}}$

Answer (iii)

$\vec{AB}\times\vec{BC}=\begin{vmatrix}\hat i&\hat j&\hat k\\-1&4&-3\\1&-3&6\end{vmatrix}=15\hat i+3\hat j-\hat k$. Area $=\frac12|\vec{AB}\times\vec{BC}|=\frac12\sqrt{225+9+1}=\frac{\sqrt{235}}{2}$ sq. units.

Final answer: $\frac{\sqrt{235}}{2}$ sq. units