Verified Solution • 2025–26 Session moderate

QJEEM17Ma10ed: JEE Main 2017 Physics - Class 11 JEEM17Ma10ed Solution: The area (in sq. units) of the parallelogram whose diagonals…

The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8i^−6j^8\widehat i - 6\widehat j and 3i^+4j^−12k^,3\widehat i + 4\widehat j - 12\widehat k, is :

Multiple Choice Options

  • A 26
  • B 65
  • C 20
  • D 52

Step-by-Step Verified Solution

When diagonal d1→{\overrightarrow {{d_1}} } and d2→{\overrightarrow {{d_2}} } are given of a parallelogram then the area of parallelogram = 12∣d1→×d2→∣{1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|

Given, d1→{\overrightarrow {{d_1}} } = 8 i^\widehat i −- 6 j^\widehat j + 0 k^\widehat k

and    d2→{\overrightarrow {{d_2}} } = 3 i^\widehat i + 4 j^\widehat j −- 12 k^\widehat k

∴   \therefore\,\,\, d1→{\overrightarrow {{d_1}} } ×\times d2→{\overrightarrow {{d_2}} }   =   |i^j^k^8−6034−12|

= 72 i^\widehat i −- ( −- 96) j^\widehat j + 50 k^\widehat k

= 72 i^\widehat i + 96 j^\widehat j + 50 k^\widehat k

∴   \therefore\,\,\, ∣d1→×d2→∣\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right| = 722+962+502\sqrt {{{72}^2} + {{96}^2} + {{50}^2}}

= 16900\sqrt {16900}

= 130

∴   \therefore\,\,\, Area of parallelogram = 12∣d1→×d2→∣{1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right| = 12{1 \over 2} ×\times 130 = 65

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