JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 1)
Let $$\mathop a\limits^ \to = 3\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge + x\mathop k\limits^ \wedge $$ and $$\mathop b\limits^ \to = \mathop i\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge $$
, for some real x. Then $$\left| {\mathop a\limits^ \to \times \mathop b\limits^ \to } \right|$$ = r
is possible if :
0 < r < $$\sqrt {{3 \over 2}} $$
$$3\sqrt {{3 \over 2}} < r < 5\sqrt {{3 \over 2}} $$
$$ r \ge 5\sqrt {{3 \over 2}} $$
$$\sqrt {{3 \over 2}} < r \le 3\sqrt {{3 \over 2}} $$
Explanation
$$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 2 & x \cr
1 & { - 1} & 1 \cr
} } \right|$$
= (2 + x)$${\widehat i}$$ + (3 - x)$${\widehat j}$$ - 5$${\widehat k}$$
$$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = r
= $$\sqrt {{{\left( {2 + x} \right)}^2} + {{\left( {x - 3} \right)}^2} + {{\left( { - 5} \right)}^2}} $$
$$ \Rightarrow $$ r = $$\sqrt {4 + 4x + {x^2} + {x^2} + 9 - 6x + 25} $$
= $$\sqrt {2{x^2} - 2x + 38} $$
= $$\sqrt {2\left( {{x^2} - x + {1 \over 4}} \right) + 38 - {1 \over 2}} $$
= $$\sqrt {2{{\left( {x - {1 \over 2}} \right)}^2} + {{75} \over 2}} $$
$$ \Rightarrow $$ r $$ \ge $$ $$\sqrt {{{75} \over 2}} $$
$$ \Rightarrow $$ $$ r \ge 5\sqrt {{3 \over 2}} $$
= (2 + x)$${\widehat i}$$ + (3 - x)$${\widehat j}$$ - 5$${\widehat k}$$
$$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = r
= $$\sqrt {{{\left( {2 + x} \right)}^2} + {{\left( {x - 3} \right)}^2} + {{\left( { - 5} \right)}^2}} $$
$$ \Rightarrow $$ r = $$\sqrt {4 + 4x + {x^2} + {x^2} + 9 - 6x + 25} $$
= $$\sqrt {2{x^2} - 2x + 38} $$
= $$\sqrt {2\left( {{x^2} - x + {1 \over 4}} \right) + 38 - {1 \over 2}} $$
= $$\sqrt {2{{\left( {x - {1 \over 2}} \right)}^2} + {{75} \over 2}} $$
$$ \Rightarrow $$ r $$ \ge $$ $$\sqrt {{{75} \over 2}} $$
$$ \Rightarrow $$ $$ r \ge 5\sqrt {{3 \over 2}} $$
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