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Let $a=p(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{c}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ be three vectors. If the value of [abc] is not more then 15 and not less than -5 , then $P$ lies in the interval
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1818 Upvotes
Verified Answer
The correct answer is:
$\left[\frac{-5}{3}, \frac{5}{9}\right]$
We have,
$\begin{aligned}
& \mathbf{a}=p \hat{\mathbf{i}}+p \hat{\mathbf{j}}+p \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \\
& \mathbf{c}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
& {[\mathbf{a} \mathbf{b} \mathbf{c}] }=\left|\begin{array}{ccc}
p & p & p \\
1 & 1 & -2 \\
2 & -1 & 2
\end{array}\right| \\
&=p(2-2)-p(2+4)+p(-1-2) \\
& \therefore[\mathbf{a b c}]=-9 p \\
& \text { Given }-5 \leq[\mathbf{a b c}] \leq 15 \\
&-5 \leq-9 p \leq 15 \\
& \therefore \quad-\frac{5}{3} \leq p \leq \frac{5}{9} \\
& p \in\left[\frac{-5}{3}, \frac{5}{9}\right]
\end{aligned}$
$\begin{aligned}
& \mathbf{a}=p \hat{\mathbf{i}}+p \hat{\mathbf{j}}+p \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \\
& \mathbf{c}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
& {[\mathbf{a} \mathbf{b} \mathbf{c}] }=\left|\begin{array}{ccc}
p & p & p \\
1 & 1 & -2 \\
2 & -1 & 2
\end{array}\right| \\
&=p(2-2)-p(2+4)+p(-1-2) \\
& \therefore[\mathbf{a b c}]=-9 p \\
& \text { Given }-5 \leq[\mathbf{a b c}] \leq 15 \\
&-5 \leq-9 p \leq 15 \\
& \therefore \quad-\frac{5}{3} \leq p \leq \frac{5}{9} \\
& p \in\left[\frac{-5}{3}, \frac{5}{9}\right]
\end{aligned}$
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