If four distinct points with position vectors a , b , c and d are coplanar, then [ a b c ] is equal to [2023]

Question

If four distinct points with position vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are coplanar, then $[\vec{a} \vec{b} \vec{c}]$ is equal to [2023]

Smart Achievers · Accepted Answer

(2)

$\vec{a}, \vec{b}, \vec{c} and \vec{d}$ are coplanar.

$∴$ $\vec{b} - \vec{a}, \vec{c} - \vec{a}, \vec{d} - \vec{a} are coplanar vectors.$

$So, [\vec{b} - \vec{a} \vec{c} - \vec{a} \vec{d} - \vec{a}] = 0$

$\Rightarrow (\vec{b} - \vec{a}) \cdot ((\vec{c} - \vec{a}) \times (\vec{d} - \vec{a})) = 0$

$\Rightarrow (\vec{b} - \vec{a}) \cdot (\vec{c} \times \vec{d} - \vec{c} \times \vec{a} - \vec{a} \times \vec{d}) = 0$

$\Rightarrow [\vec{b} \vec{c} \vec{d}] - [\vec{b} \vec{c} \vec{a}] - [\vec{b} \vec{a} \vec{d}] - [\vec{a} \vec{c} \vec{d}] = 0$

$∴$ $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{d}] - [\vec{b} \vec{a} \vec{d}] - [\vec{a} \vec{c} \vec{d}]$

$= [\vec{c} \vec{d} \vec{b}] + [\vec{b} \vec{d} \vec{a}] + [\vec{d} \vec{c} \vec{a}]$

Solution

Q.
If four distinct points with position vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are coplanar, then $[\vec{a} \vec{b} \vec{c}]$ is equal to [2023]

1 $[\vec{b} \vec{c} \vec{d}] + [\vec{d} \vec{a} \vec{c}] + [\vec{d} \vec{b} \vec{a}]$

2 $[\vec{d} \vec{c} \vec{a}] + [\vec{b} \vec{d} \vec{a}] + [\vec{c} \vec{d} \vec{b}]$

3 $[\vec{d} \vec{b} \vec{a}] + [\vec{a} \vec{c} \vec{d}] + [\vec{d} \vec{b} \vec{c}]$

4 $[\vec{a} \vec{d} \vec{b}] + [\vec{d} \vec{c} \vec{a}] + [\vec{d} \vec{b} \vec{c}]$

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Solution

Q. If four distinct points with position vectors a→,b→,c→ and d→ are coplanar, then [a→b→c→] is equal to [2023]

1 [b→c→d→]+[d→a→c→]+[d→b→a→]

2 [d→c→a→]+[b→d→a→]+[c→d→b→]

3 [d→b→a→]+[a→c→d→]+[d→b→c→]