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CBSE Sample Paper 2020-21 · Part B: Section III · 2 marks

Q26: Parallelogram with side i^−j^+k^\hat i - \hat j + \hat k and diagonal 4i^+5k^4\hat i + 5\hat k⁠: its area

Sample 2020-21Ch 10 · Vector Algebra
The question

Find the area of the parallelogram whose one side and a diagonal are represented by coinitial vectors i^−j^+k^\hat i - \hat j + \hat k and 4i^+5k^4\hat i + 5\hat k respectively

Exam answer

Let a⃗=i^−j^+k^\vec a = \hat i - \hat j + \hat k (a side) and d⃗=4i^+5k^\vec d = 4\hat i + 5\hat k (the diagonal). If b⃗\vec b is the other side, then a⃗+b⃗=d⃗\vec a + \vec b = \vec d⁠.

Why?

The side and the diagonal start from the same point (coinitial). By the parallelogram law, the diagonal from that point is the sum of the two sides. Watch out: it is a⃗+b⃗\vec a + \vec b⁠, not a⃗−b⃗\vec a - \vec b⁠.

∴ b⃗=d⃗−a⃗=3i^+j^+4k^\begin{aligned} \therefore\ \vec b &= \vec d - \vec a \\ &= 3\hat i + \hat j + 4\hat k \end{aligned}
Why?

Subtract component by component: (4−1)i^+(0+1)j^+(5−1)k^(4 - 1)\hat i + (0 + 1)\hat j + (5 - 1)\hat k⁠. The diagonal has no j^\hat j part, so its j^\hat j component is 0.

b⃗=(4i^+0j^+5k^)−(i^−j^+k^)=(4−1)i^+(0−(−1))j^+(5−1)k^=3i^+1j^+4k^\begin{aligned} \vec b &= (4\hat i + 0\hat j + 5\hat k) \\ &\quad - (\hat i - \hat j + \hat k) \\ &= (4 - 1)\hat i + (0 - (-1))\hat j \\ &\quad + (5 - 1)\hat k \\ &= 3\hat i + 1\hat j + 4\hat k \end{aligned}
a⃗×b⃗=∣i^j^k^1−11314∣=i^(−4−1)−j^(4−3)+k^(1+3)=−5i^−j^+4k^\begin{aligned} \vec a \times \vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 1 \\ 3 & 1 & 4 \end{vmatrix} \\ &= \hat i(-4 - 1) - \hat j(4 - 3) \\ &\quad + \hat k(1 + 3) \\ &= -5\hat i - \hat j + 4\hat k \end{aligned}
Why?

Expand the determinant along the first row:

i^(a2b3−a3b2)−j^(a1b3−a3b1)+k^(a1b2−a2b1)\begin{aligned} &\hat i(a_2b_3 - a_3b_2) \\ &- \hat j(a_1b_3 - a_3b_1) \\ &+ \hat k(a_1b_2 - a_2b_1) \end{aligned}

For example, the i^\hat i part is (−1)(4)−(1)(1)=−5(-1)(4) - (1)(1) = -5⁠. Watch out for the minus sign in front of the j^\hat j term.

Cover the i^\hat i column, then the j^\hat j column, then the k^\hat k column:

i^:∣−1114∣=(−1)(4)−(1)(1)=−4−1=−5j^:∣1134∣=(1)(4)−(1)(3)=4−3=1with the minus: −1k^:∣1−131∣=(1)(1)−(−1)(3)=1+3=4\begin{aligned} \hat i&: \begin{vmatrix} -1 & 1 \\ 1 & 4 \end{vmatrix} \\ &= (-1)(4) - (1)(1) \\ &= -4 - 1 = -5 \\[4pt] \hat j&: \begin{vmatrix} 1 & 1 \\ 3 & 4 \end{vmatrix} \\ &= (1)(4) - (1)(3) \\ &= 4 - 3 = 1 \\ &\text{with the minus: } -1 \\[4pt] \hat k&: \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} \\ &= (1)(1) - (-1)(3) \\ &= 1 + 3 = 4 \end{aligned}
Area=∣a⃗×b⃗∣=(−5)2+(−1)2+42=25+1+16\begin{aligned} \text{Area} &= |\vec a \times \vec b| \\ &= \sqrt{(-5)^2 + (-1)^2 + 4^2} \\ &= \sqrt{25 + 1 + 16} \end{aligned}
Why?

The area of a parallelogram with adjacent sides a⃗\vec a and b⃗\vec b is ∣a⃗×b⃗∣|\vec a \times \vec b|⁠. The magnitude of xi^+yj^+zk^x\hat i + y\hat j + z\hat k is x2+y2+z2\sqrt{x^2 + y^2 + z^2}⁠.

(−5)2=25, (−1)2=1, 42=1625+1=2626+16=42\begin{aligned} (-5)^2 &= 25,\ (-1)^2 = 1,\ 4^2 = 16 \\ 25 + 1 &= 26 \\ 26 + 16 &= 42 \end{aligned}

So the area is 42\sqrt{42}⁠. 42 has no square factor, so 42\sqrt{42} does not simplify.

∴\therefore Area of the parallelogram is 42\sqrt{42} sq. units

Where this comes from

Chapter 10, Vector Algebra: Area of a parallelogram by the cross product.

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, the other side is diagonal minus side, b⃗=3i^+j^+4k^\vec b = 3\hat i + \hat j + 4\hat k⁠. Then the area is ∣a⃗×b⃗∣|\vec a \times \vec b|⁠, from the determinant.

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