Question
Find the value of so $\lambda$ that the lines
$\frac{1 - x}{3} = \frac{7y - 14}{2\lambda} = \frac{5z - 10}{11} \text{and} \frac{7 - 7x}{3\lambda} = \frac{y - 5}{1}= \frac{6 - z}{5}$
are perpendicular to each other.
$\frac{1 - x}{3} = \frac{7y - 14}{2\lambda} = \frac{5z - 10}{11} \text{and} \frac{7 - 7x}{3\lambda} = \frac{y - 5}{1}= \frac{6 - z}{5}$
are perpendicular to each other.
$(a_{1}, b_{1}, c_{1},) = \bigg( -3, \frac{2\lambda}{7}, \frac{11}{5}\bigg) \text{and} (a_{2}, b_{2}, c_{2},) = \bigg(-\frac{3\lambda}{7},1, -5\bigg)$
Two lines are perpendicular then $a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0$
$\Rightarrow \frac{9\lambda}{7} + \frac{2\lambda}{7} - 11 = 0 \Rightarrow \frac{11\lambda}{7} = 11 \Rightarrow \lambda = 7$
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Solve the following LPP graphically:
Maximize Z = 20x + 10y
Subject to the following constraints
$\text{x}+2\text{y}\leq28$
$3\text{x}+\text{y}\leq24$
$\text{x}\geq2$
$\text{x},\text{y}\geq0$