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Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsVectors4 Marks
Question
If four points $\mathrm{A}(\bar{a}), \mathrm{B}(\bar{b}), C(\bar{c})$ and $\mathrm{D}(\bar{d})$ are coplanar then show that

$\left[\begin{array}{lll}\bar{a} \bar{b} \bar{d}\end{array}\right]+\left[\begin{array}{lll}\bar{b} & \bar{c} & \bar{d}\end{array}\right]+\left[\begin{array}{lll}\bar{c} & \bar{a} & \bar{d}\end{array}\right]=\left[\begin{array}{ll}\bar{a} \bar{b} & \bar{c}\end{array}\right]$

✓

Answer

$a, \bar{b}, \bar{c}$ and $\bar{d}$ are the position vectors of the

points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ arid $\mathrm{D}$ respectively.

$\therefore \overline{\mathrm{AB}}=\bar{b}-\bar{a}, \overline{\mathrm{AC}}=\overline{\mathrm{c}}-\bar{a}, \overline{\mathrm{AD}}=\bar{d}-\bar{a}$

The points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are coplanar.

$\therefore$ the vectors $\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}$ are woplanar.

$\begin{aligned} & \therefore[\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]=0 \\ & \therefore[\bar{b}-\bar{a} \bar{c}-\bar{a} \bar{d}-\bar{a}]=0 \\ & \therefore(\bar{b}-\bar{a}) \cdot[(\bar{c}-\bar{a}) \times(\bar{d}-\bar{a})]=0 \\ & \therefore(\bar{b}-\bar{a}) \cdot(\bar{c} \times \bar{d}-\bar{c} \times \bar{a}-\bar{a} \times \bar{d}+\bar{a} \times \bar{a})=0\end{aligned}$

where $\bar{a} \times \bar{a}=\overline{0}$

$\therefore(\bar{b}-\bar{a}) \cdot(\bar{c} \times \bar{d}-\bar{c} \times \bar{a}-\bar{a} \times \bar{d})=0$

$\begin{array}{r}\therefore \bar{b} \cdot(\bar{c} \times \bar{d})-\bar{b} \cdot(\bar{c} \times \bar{a})-\bar{b} \cdot(\bar{a} \times \bar{d})-\bar{a} \cdot(\bar{c} \times \bar{d})+ \\ \bar{a} \cdot(\bar{c} \times \bar{a})+\bar{a} \cdot(\bar{a} \times \bar{d})=0 \quad \ldots(1)\end{array}$

Now, $\bar{a} \cdot(\bar{c} \times \bar{a})=0, \bar{a} \cdot(\bar{a} \times \bar{d})=0$,

$\begin{aligned} & -\bar{b} \cdot(\bar{a} \times \bar{d})=\bar{b} \cdot(\bar{d} \times \bar{a})=[\bar{b} \bar{d} \bar{a}]=[\bar{a} \bar{b} \bar{d}], \\ & -\bar{a} \cdot(\bar{c} \times \bar{d})=\bar{a} \cdot(\bar{d} \times \bar{c})=[\bar{a} \bar{d} \bar{c}]=[\bar{c} \bar{a} \bar{d}]\end{aligned}$

Also, $\bar{b} \cdot(\bar{c} \times \bar{d})=[\bar{b} \bar{c} \bar{d}]$,

$-\bar{b} \cdot(\bar{c} \times \bar{a})=-\bar{a} \cdot(\bar{b} \times \bar{c})=-[\bar{a} \bar{b} \bar{c}]$

$\therefore$ from (1)

$\begin{aligned} & {[\bar{b} \bar{c} \bar{d}]-[\bar{a} \bar{b} \bar{c}]+[\bar{a} \bar{b} \bar{d}]+[\bar{c} \bar{a} \bar{d}]+0+0=0} \\ & \therefore[\bar{a} \bar{b} \bar{d}]+[\bar{b} \bar{c} \bar{d}]+[\bar{c} \bar{a} \bar{d}]=[\bar{a} \bar{b} \bar{c}] .\end{aligned}$

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