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CBSE Board Examination 2022 (Term 2) · Section A · 2 marks

Q2: Diagonals of parallelogram PQRS with PQ→=3i^−2j^+2k^\overrightarrow{PQ} = 3\hat i - 2\hat j + 2\hat k⁠, PS→=−i^−2k^\overrightarrow{PS} = -\hat i - 2\hat k

Board 2022 T2Ch 10 · Vector Algebra
The question

In a parallelogram PQRS, PQ→=3i^−2j^+2k^\overrightarrow{PQ} = 3\hat i - 2\hat j + 2\hat k and PS→=−i^−2k^\overrightarrow{PS} = -\hat i - 2\hat k⁠. Find ∣PR→∣|\overrightarrow{PR}| and ∣QS→∣|\overrightarrow{QS}|⁠.

Exam answer

In parallelogram PQRS, QR→=PS→\overrightarrow{QR} = \overrightarrow{PS}⁠.

Why?

Opposite sides of a parallelogram are equal in length and parallel, and QR and PS point the same way. Two vectors with the same length and the same direction are equal.

PR→=PQ→+QR→=PQ→+PS→=(3−1)i^−2j^+(2−2)k^=2i^−2j^\begin{aligned} \overrightarrow{PR} &= \overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PQ} + \overrightarrow{PS} \\ &= (3 - 1)\hat i - 2\hat j + (2 - 2)\hat k \\ &= 2\hat i - 2\hat j \end{aligned}
Why?

Triangle law: going from P to Q and then from Q to R is the same as going straight from P to R.

Then add matching components. PS→\overrightarrow{PS} has no j^\hat j part, so its j^\hat j component is 0:

i^: 3+(−1)=2j^: −2+0=−2k^: 2+(−2)=0\begin{aligned} \hat i&:\ 3 + (-1) = 2 \\ \hat j&:\ -2 + 0 = -2 \\ \hat k&:\ 2 + (-2) = 0 \end{aligned}
∣PR→∣=4+4=22|\overrightarrow{PR}| = \sqrt{4 + 4} = 2\sqrt2
Why?

The length (magnitude) of a vector is

∣ai^+bj^+ck^∣=a2+b2+c2\begin{aligned} &|a\hat i + b\hat j + c\hat k| \\ &= \sqrt{a^2 + b^2 + c^2} \end{aligned}

Here a=2a = 2⁠, b=−2b = -2⁠, c=0c = 0⁠:

∣PR→∣=(2)2+(−2)2+(0)2=4+4+0=8=4×2=22\begin{aligned} |\overrightarrow{PR}| &= \sqrt{(2)^2 + (-2)^2 + (0)^2} \\ &= \sqrt{4 + 4 + 0} \\ &= \sqrt8 \\ &= \sqrt{4 \times 2} = 2\sqrt2 \end{aligned}
QS→=QP→+PS→=PS→−PQ→=(−1−3)i^+2j^+(−2−2)k^=−4i^+2j^−4k^\begin{aligned} \overrightarrow{QS} &= \overrightarrow{QP} + \overrightarrow{PS} = \overrightarrow{PS} - \overrightarrow{PQ} \\ &= (-1 - 3)\hat i + 2\hat j + (-2 - 2)\hat k \\ &= -4\hat i + 2\hat j - 4\hat k \end{aligned}
Why?

Go from Q back to P, then from P to S. Going backwards along PQ reverses the vector, so QP→=−PQ→\overrightarrow{QP} = -\overrightarrow{PQ}⁠.

Subtract matching components. PS→\overrightarrow{PS} has j^\hat j component 0:

i^: −1−3=−4j^: 0−(−2)=0+2=2k^: −2−2=−4\begin{aligned} \hat i&:\ -1 - 3 = -4 \\ \hat j&:\ 0 - (-2) = 0 + 2 = 2 \\ \hat k&:\ -2 - 2 = -4 \end{aligned}

Watch the j^\hat j part: 0−(−2)=+20 - (-2) = +2⁠.

The length then uses the same formula:

∣QS→∣=(−4)2+(2)2+(−4)2=16+4+16=36=6\begin{aligned} |\overrightarrow{QS}| &= \sqrt{(-4)^2 + (2)^2 + (-4)^2} \\ &= \sqrt{16 + 4 + 16} \\ &= \sqrt{36} = 6 \end{aligned}
∣QS→∣=16+4+16=6|\overrightarrow{QS}| = \sqrt{16 + 4 + 16} = 6

∴ ∣PR→∣=22\therefore\ |\overrightarrow{PR}| = 2\sqrt2 and ∣QS→∣=6|\overrightarrow{QS}| = 6⁠.

Where this comes from

Chapter 10, Vector Algebra: Diagonals of a parallelogram from its sides.

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