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CBSE Sample Paper 2024-25 · Section B · 2 marks

Q25: Diagonals and area of the parallelogram on 2i^−4j^−5k^2\hat i - 4\hat j - 5\hat k and 2i^+2j^+3k^2\hat i + 2\hat j + 3\hat k

Sample 2024-25Ch 10 · Vector Algebra
The question

The two co-initial adjacent sides of a parallelogram are 2i^−4j^−5k^2\hat i - 4\hat j - 5\hat k and 2i^+2j^+3k^2\hat i + 2\hat j + 3\hat k⁠. Find its diagonals and use them to find the area of the parallelogram.

Exam answer

Let a⃗=2i^−4j^−5k^\vec a = 2\hat i - 4\hat j - 5\hat k and b⃗=2i^+2j^+3k^\vec b = 2\hat i + 2\hat j + 3\hat k⁠. The diagonals are

d1⃗=a⃗+b⃗=4i^−2j^−2k^d2⃗=a⃗−b⃗=−6j^−8k^\begin{aligned} \vec{d_1} &= \vec a + \vec b = 4\hat i - 2\hat j - 2\hat k \\ \vec{d_2} &= \vec a - \vec b = -6\hat j - 8\hat k \end{aligned}
Why?

Place a⃗\vec a and b⃗\vec b along two adjacent sides from the same corner. By the triangle law, one diagonal is a⃗+b⃗\vec a + \vec b⁠.

The other diagonal joins the tips of b⃗\vec b and a⃗\vec a⁠. It is a⃗−b⃗\vec a - \vec b⁠, because b⃗+(a⃗−b⃗)=a⃗\vec b + (\vec a - \vec b) = \vec a⁠.

Add or subtract like components. In a⃗−b⃗\vec a - \vec b the i^\hat i parts cancel (2−2=02 - 2 = 0⁠). Watch the signs: −4−2=−6-4 - 2 = -6 and −5−3=−8-5 - 3 = -8⁠.

a⃗+b⃗:i^:2+2=4j^:−4+2=−2k^:−5+3=−2a⃗−b⃗:i^:2−2=0j^:−4−2=−6k^:−5−3=−8\begin{aligned} \vec a + \vec b&: \\ \hat i&: 2 + 2 = 4 \\ \hat j&: -4 + 2 = -2 \\ \hat k&: -5 + 3 = -2 \\[4pt] \vec a - \vec b&: \\ \hat i&: 2 - 2 = 0 \\ \hat j&: -4 - 2 = -6 \\ \hat k&: -5 - 3 = -8 \end{aligned}
d1⃗×d2⃗=∣i^j^k^4−2−20−6−8∣=i^(16−12)−j^(−32−0)+k^(−24−0)=4i^+32j^−24k^\begin{aligned} \vec{d_1} \times \vec{d_2} &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & -2 & -2 \\ 0 & -6 & -8 \end{vmatrix} \\ &= \hat i(16 - 12) \\ &\quad - \hat j(-32 - 0) \\ &\quad + \hat k(-24 - 0) \\ &= 4\hat i + 32\hat j - 24\hat k \end{aligned}
Why?

Expand the determinant along the first row with the signs ++⁠, −-⁠, ++⁠. Each unit vector is multiplied by the 2×22 \times 2 determinant left when its row and column are covered:

i^:(−2)(−8)−(−2)(−6)j^:(4)(−8)−(−2)(0)k^:(4)(−6)−(−2)(0)\begin{aligned} \hat i&: (-2)(-8) - (-2)(-6) \\ \hat j&: (4)(-8) - (-2)(0) \\ \hat k&: (4)(-6) - (-2)(0) \end{aligned}

Work out each one:

i^:∣−2−2−6−8∣=(−2)(−8)−(−2)(−6)=16−12=4j^:∣4−20−8∣=(4)(−8)−(−2)(0)=−32−0=−32k^:∣4−20−6∣=(4)(−6)−(−2)(0)=−24−0=−24\begin{aligned} \hat i&: \begin{vmatrix} -2 & -2 \\ -6 & -8 \end{vmatrix} \\ &= (-2)(-8) - (-2)(-6) \\ &= 16 - 12 = 4 \\[4pt] \hat j&: \begin{vmatrix} 4 & -2 \\ 0 & -8 \end{vmatrix} \\ &= (4)(-8) - (-2)(0) \\ &= -32 - 0 = -32 \\[4pt] \hat k&: \begin{vmatrix} 4 & -2 \\ 0 & -6 \end{vmatrix} \\ &= (4)(-6) - (-2)(0) \\ &= -24 - 0 = -24 \end{aligned}

Now apply the signs ++⁠, −-⁠, ++⁠:

+i^(4)=4i^−j^(−32)=+32j^+k^(−24)=−24k^\begin{aligned} +\hat i(4) &= 4\hat i \\ -\hat j(-32) &= +32\hat j \\ +\hat k(-24) &= -24\hat k \end{aligned}

Watch out: the j^\hat j term carries a minus sign, so −j^(−32)=+32j^-\hat j(-32) = +32\hat j⁠.

Area=12∣d1⃗×d2⃗∣=2∣i^+8j^−6k^∣=21+64+36\begin{aligned} \text{Area} &= \frac12\lvert\vec{d_1} \times \vec{d_2}\rvert \\ &= 2\lvert\hat i + 8\hat j - 6\hat k\rvert \\ &= 2\sqrt{1 + 64 + 36} \end{aligned}
Why?

If d1⃗\vec{d_1} and d2⃗\vec{d_2} are the diagonals of a parallelogram, its area is half the magnitude of their cross product.

Take out the common factor 4, and use 12×4=2\frac12 \times 4 = 2⁠:

4i^+32j^−24k^=4(i^+8j^−6k^)\begin{aligned} &4\hat i + 32\hat j - 24\hat k \\ &= 4(\hat i + 8\hat j - 6\hat k) \end{aligned} 12∣4(i^+8j^−6k^)∣=12×4 ∣i^+8j^−6k^∣=2 ∣i^+8j^−6k^∣\begin{aligned} &\frac12\lvert 4(\hat i + 8\hat j - 6\hat k)\rvert \\ &= \frac12 \times 4\,\lvert\hat i + 8\hat j - 6\hat k\rvert \\ &= 2\,\lvert\hat i + 8\hat j - 6\hat k\rvert \end{aligned}

Then 12+82+(−6)2=1+64+36=1011^2 + 8^2 + (-6)^2 = 1 + 64 + 36 = 101⁠.

1+64=6565+36=101\begin{aligned} 1 + 64 &= 65 \\ 65 + 36 &= 101 \end{aligned}

So the area is 21012\sqrt{101}⁠.

Check: the sides give the same area, since ∣a⃗×b⃗∣=404=2101\lvert\vec a \times \vec b\rvert = \sqrt{404} = 2\sqrt{101}⁠.

a⃗×b⃗=∣i^j^k^2−4−5223∣i^:(−4)(3)−(−5)(2)=−12+10=−2j^:−((2)(3)−(−5)(2))=−(6+10)=−16k^:(2)(2)−(−4)(2)=4+8=12∣a⃗×b⃗∣2=(−2)2+(−16)2+122=4+256+144=404∣a⃗×b⃗∣=404=4×101=2101\begin{aligned} \vec a \times \vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -4 & -5 \\ 2 & 2 & 3 \end{vmatrix} \\[4pt] \hat i&: (-4)(3) - (-5)(2) \\ &= -12 + 10 = -2 \\ \hat j&: -\big((2)(3) - (-5)(2)\big) \\ &= -(6 + 10) = -16 \\ \hat k&: (2)(2) - (-4)(2) \\ &= 4 + 8 = 12 \\[4pt] \lvert\vec a \times \vec b\rvert^2 &= (-2)^2 + (-16)^2 + 12^2 \\ &= 4 + 256 + 144 \\ &= 404 \\ \lvert\vec a \times \vec b\rvert &= \sqrt{404} = \sqrt{4 \times 101} \\ &= 2\sqrt{101} \end{aligned}

∴\therefore Area of the parallelogram =2101= 2\sqrt{101} sq. units.

Where this comes from

Chapter 10, Vector Algebra: Area of a parallelogram from its diagonals.

Question type: Cross product, and the area of a parallelogram

How to spot it: click to see

Type 8. Cross product, and the area of a parallelogram

Spot it: "find a⃗×b⃗\vec a\times\vec b", "find ∣a⃗×b⃗∣|\vec a\times\vec b|", or "the area of the parallelogram whose adjacent sides are".

Here: First write the diagonals a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b⁠. Their cross product is 4i^+32j^−24k^4\hat i + 32\hat j - 24\hat k⁠, and the area is half its length, because they are diagonals.

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