JEE MAIN - Mathematics (2017 - 8th April Morning Slot - No. 15)
The area (in sq. units) of the parallelogram whose diagonals are along the vectors $$8\widehat i - 6\widehat j$$ and $$3\widehat i + 4\widehat j - 12\widehat k,$$ is :
26
65
20
52
Explanation
When diagonal $${\overrightarrow {{d_1}} }$$ and $${\overrightarrow {{d_2}} }$$ are given of a parallelogram then the area of parallelogram = $${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$
Given, $${\overrightarrow {{d_1}} }$$ = 8$$\widehat i$$ $$-$$ 6$$\widehat j$$ + 0$$\widehat k$$
and $${\overrightarrow {{d_2}} }$$ = 3$$\widehat i$$ + 4$$\widehat j$$ $$-$$ 12$$\widehat k$$
$$\therefore\,\,\,$$ $${\overrightarrow {{d_1}} }$$ $$ \times $$ $${\overrightarrow {{d_2}} }$$ = $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 8 & { - 6} & 0 \cr 3 & 4 & { - 12} \cr } } \right|$$
= 72 $$\widehat i$$ $$-$$ ($$-$$ 96) $$\widehat j$$ + 50$$\widehat k$$
= 72 $$\widehat i$$ + 96 $$\widehat j$$ + 50 $$\widehat k$$
$$\therefore\,\,\,$$ $$\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$ = $$\sqrt {{{72}^2} + {{96}^2} + {{50}^2}} $$
= $$\sqrt {16900} $$
= 130
$$\therefore\,\,\,$$ Area of parallelogram = $${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$ = $${1 \over 2}$$ $$ \times $$ 130 = 65
Given, $${\overrightarrow {{d_1}} }$$ = 8$$\widehat i$$ $$-$$ 6$$\widehat j$$ + 0$$\widehat k$$
and $${\overrightarrow {{d_2}} }$$ = 3$$\widehat i$$ + 4$$\widehat j$$ $$-$$ 12$$\widehat k$$
$$\therefore\,\,\,$$ $${\overrightarrow {{d_1}} }$$ $$ \times $$ $${\overrightarrow {{d_2}} }$$ = $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 8 & { - 6} & 0 \cr 3 & 4 & { - 12} \cr } } \right|$$
= 72 $$\widehat i$$ $$-$$ ($$-$$ 96) $$\widehat j$$ + 50$$\widehat k$$
= 72 $$\widehat i$$ + 96 $$\widehat j$$ + 50 $$\widehat k$$
$$\therefore\,\,\,$$ $$\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$ = $$\sqrt {{{72}^2} + {{96}^2} + {{50}^2}} $$
= $$\sqrt {16900} $$
= 130
$$\therefore\,\,\,$$ Area of parallelogram = $${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$ = $${1 \over 2}$$ $$ \times $$ 130 = 65
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