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Scalar Triple Product — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsScalar Triple Product5 Marks
Question
If the vectors $\big(\sec^2\text{A}\big)\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\hat{\text{i}}+\big(\sec^2\text{B}\big)+\hat{\text{k}},\hat{\text{i}}+\hat{\text{j}}+\big(\sec^2\text{C}\big)\hat{\text{k}}$ are coplanar, then find the value of $\text{cosec}^2\text{A}+\text{cosec}^2\text{B}+\text{cosec}^2\text{C}.$
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Answer

Let: $\vec{\text{a}}=\big(\sec^2\text{A}\big)\hat{\text{i}}+\hat{\text{j}},\vec{\text{b}}=\hat{\text{i}}+(\sec^2\text{B})\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+(\sec^2\text{C})\hat{\text{k}}$
We know that three vectors are coplanar if their scaler triple product is zero i.e., $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
Here, $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
$\begin{vmatrix}\sec^2\text{A}&1&1\\1&\sec^2\text{B}&1\\1&1&\sec^2\text{C} \end{vmatrix}=0$
$\Rightarrow\sec^2\text{A}\big[\big(\sec^2\text{B}\times\sec^2\text{C}\big)\\-1\big]-1\big(\sec^2\text{C}-1\big)+1\big(1-\sec^2\text{B}\big)=0$
$\Rightarrow\sec^2\text{A}\sec^2\text{B}\sec^2\text{C}-\sec^2\text{A}-\sec^2\text{C}+1+1-\sec^2\text{B}=0$
$\Rightarrow\big(1+\tan^2\text{A}\big)\big(1+\tan^2\text{B}\big)\big(1+\tan^2\text{C}\big)\\-\big(1+\tan^2\text{A}\big)-\big(1+\tan^2\text{C}\big)+1=1-\big(1+\tan^2\text{B}\big)=0$
$\Rightarrow1+\tan^2\text{A}+\tan^2\text{B}+\tan^2\text{C}+\tan^2\text{A}\tan^2\text{B}\\+\tan^\text{B}\tan^2\text{C}+\tan^2\text{C}\tan^2\text{A}+\tan^2\text{A}\tan^2\text{B}\tan^\text{C}1\\-\tan^2\text{A}-1-\tan^2\text{C}$
$\tan^2\text{A}\tan^2\text{B}+\tan^2\text{B}\tan^2\text{C}+\tan^2\text{C}\tan^2\text{A}\\+\tan^2\text{A}\tan^2\text{B}\tan^2\text{C}=0$
$\Rightarrow\tan^2\text{A}\tan^2\text{B}+\tan^2\text{B}\tan^2\text{C}+\tan^2\text{C}\tan^2\text{A}\\=-\tan^2\text{A}\tan^2\text{B}\tan^2\text{C}$
$\Rightarrow\frac{\tan^2\text{A}\tan^2\text{B}\tan^2\text{C}+\tan^2\text{C}\tan^2\text{A}}{\tan^2\text{A}\tan^2\text{B}\tan^2\text{C}}=-1$
$\Rightarrow\cot^2\text{C}+\cot^2\text{A}+\cot^2\text{B}=-1$
$\Rightarrow\text{cosec}^2\text{C}-1+\text{cosec}^2\text{A}-1+\text{cosec}^2\text{B}-1=-1$
$\therefore\text{cosec}^2\text{A}+\text{cosec}^2\text{B}+\text{cosec}^2\text{C}=2$

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