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Q.Find the area of the parallelogram whose diagonals are 3î + ĵ − 2k̂ and î − 3ĵ + 4k̂. Also find its adjacent sides.

Goa GbshseGBSHSE Class 12 Board Exam 2025Subjective· 3mImportance★★★★★
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For a parallelogram with diagonal vectors d1⃗,d2⃗\vec{d_1}, \vec{d_2}, the area is 12∣d1⃗×d2⃗∣\tfrac12|\vec{d_1}\times\vec{d_2}|, and the adjacent sides are recovered from a⃗=12(d1⃗+d2⃗)\vec a=\tfrac12(\vec{d_1}+\vec{d_2}), b⃗=12(d1⃗−d2⃗)\vec b=\tfrac12(\vec{d_1}-\vec{d_2}) (since the diagonals of a parallelogram with sides a⃗,b⃗\vec a,\vec b are a⃗+b⃗\vec a+\vec b and a⃗−b⃗\vec a-\vec b).

Given: diagonals d1⃗=3i^+j^−2k^\vec{d_1} = 3\hat i+\hat j-2\hat k, d2⃗=i^−3j^+4k^\vec{d_2} = \hat i-3\hat j+4\hat k

Step 1 — cross product d1⃗×d2⃗\vec{d_1}\times\vec{d_2}:

d1⃗×d2⃗=∣i^j^k^31−21−34∣\vec{d_1}\times\vec{d_2} = \begin{vmatrix}\hat i & \hat j & \hat k\\ 3 & 1 & -2\\ 1 & -3 & 4\end{vmatrix}

=i^(1⋅4−(−2)(−3))−j^(3⋅4−(−2)⋅1)+k^(3⋅(−3)−1⋅1)= \hat i(1\cdot4-(-2)(-3)) - \hat j(3\cdot4-(-2)\cdot1) + \hat k(3\cdot(-3)-1\cdot1)

=i^(4−6)−j^(12+2)+k^(−9−1)=−2i^−14j^−10k^= \hat i(4-6) - \hat j(12+2) + \hat k(-9-1) = -2\hat i - 14\hat j - 10\hat k

Step 2 — magnitude:

∣d1⃗×d2⃗∣=(−2)2+(−14)2+(−10)2=4+196+100=300=103|\vec{d_1}\times\vec{d_2}| = \sqrt{(-2)^2+(-14)^2+(-10)^2} = \sqrt{4+196+100} = \sqrt{300} = 10\sqrt3

Step 3 — area of parallelogram:

Area=12∣d1⃗×d2⃗∣=12(103)=53 sq. units\text{Area} = \frac12|\vec{d_1}\times\vec{d_2}| = \frac12(10\sqrt3) = 5\sqrt3 \text{ sq. units}

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