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CBSE Board Examination 2022 (Term 2) · Section B · 3 marks

Q7: Show (a⃗−2d⃗)∥(2b⃗−c⃗)(\vec a - 2\vec d) \parallel (2\vec b - \vec c)⁠, or unit vectors along the diagonals and the area

Board 2022 T2Ch 10 · Vector Algebra
The question

(a) If a⃗\vec a⁠, b⃗\vec b⁠, c⃗\vec c and d⃗\vec d are four non-zero vectors such that a⃗×b⃗=c⃗×d⃗\vec a \times \vec b = \vec c \times \vec d and a⃗×c⃗=4b⃗×d⃗\vec a \times \vec c = 4\vec b \times \vec d⁠, then show that (a⃗−2d⃗)(\vec a - 2\vec d) is parallel to (2b⃗−c⃗)(2\vec b - \vec c) where a⃗≠2d⃗\vec a \ne 2\vec d⁠, c⃗≠2b⃗\vec c \ne 2\vec b⁠.

OR

(b) The two adjacent sides of a parallelogram are represented by 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 the unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram also.

Exam answer

(a)

Consider (a⃗−2d⃗)×(2b⃗−c⃗)(\vec a - 2\vec d) \times (2\vec b - \vec c)⁠.

Why?

Two non-zero vectors are parallel exactly when their cross product is the zero vector, because

∣u⃗×v⃗∣=∣u⃗∣∣v⃗∣sin⁡θ|\vec u \times \vec v| = |\vec u||\vec v|\sin\theta

and sin⁡θ=0\sin\theta = 0 only for θ=0\theta = 0 or π\pi⁠. So the plan is to show that this cross product is 0⃗\vec 0⁠.

(a⃗−2d⃗)×(2b⃗−c⃗)=2 a⃗×b⃗−a⃗×c⃗−4 d⃗×b⃗+2 d⃗×c⃗\begin{aligned} &(\vec a - 2\vec d) \times (2\vec b - \vec c) \\ &= 2\,\vec a \times \vec b - \vec a \times \vec c \\ &\quad - 4\,\vec d \times \vec b + 2\,\vec d \times \vec c \end{aligned}
Why?

The cross product distributes over addition, so multiply out the brackets like ordinary brackets, keeping the order inside each pair:

a⃗×2b⃗=2 a⃗×b⃗a⃗×(−c⃗)=−a⃗×c⃗(−2d⃗)×2b⃗=−4 d⃗×b⃗(−2d⃗)×(−c⃗)=2 d⃗×c⃗\begin{aligned} \vec a \times 2\vec b &= 2\,\vec a \times \vec b \\ \vec a \times (-\vec c) &= -\vec a \times \vec c \\ (-2\vec d) \times 2\vec b &= -4\,\vec d \times \vec b \\ (-2\vec d) \times (-\vec c) &= 2\,\vec d \times \vec c \end{aligned}

Watch out: never swap the two vectors in a pair here. The cross product is not commutative.

Using d⃗×b⃗=−b⃗×d⃗\vec d \times \vec b = -\vec b \times \vec d and d⃗×c⃗=−c⃗×d⃗\vec d \times \vec c = -\vec c \times \vec d⁠,

=2 a⃗×b⃗−a⃗×c⃗+4 b⃗×d⃗−2 c⃗×d⃗\begin{aligned} &= 2\,\vec a \times \vec b - \vec a \times \vec c \\ &\quad + 4\,\vec b \times \vec d - 2\,\vec c \times \vec d \end{aligned}
Why?

Swapping the order of a cross product changes its sign: u⃗×v⃗=−v⃗×u⃗\vec u \times \vec v = -\vec v \times \vec u⁠.

The given conditions use b⃗×d⃗\vec b \times \vec d and c⃗×d⃗\vec c \times \vec d⁠, so we swap to make the terms look like them.

−4 d⃗×b⃗=−4(−b⃗×d⃗)=+4 b⃗×d⃗2 d⃗×c⃗=2(−c⃗×d⃗)=−2 c⃗×d⃗\begin{aligned} &-4\,\vec d \times \vec b \\ &= -4(-\vec b \times \vec d) \\ &= +4\,\vec b \times \vec d \\[4pt] &2\,\vec d \times \vec c \\ &= 2(-\vec c \times \vec d) \\ &= -2\,\vec c \times \vec d \end{aligned}

The first two terms do not change.

Putting a⃗×b⃗=c⃗×d⃗\vec a \times \vec b = \vec c \times \vec d and a⃗×c⃗=4b⃗×d⃗\vec a \times \vec c = 4\vec b \times \vec d⁠,

=2 c⃗×d⃗−4 b⃗×d⃗+4 b⃗×d⃗−2 c⃗×d⃗=0⃗\begin{aligned} &= 2\,\vec c \times \vec d - 4\,\vec b \times \vec d \\ &\quad + 4\,\vec b \times \vec d - 2\,\vec c \times \vec d \\ &= \vec 0 \end{aligned}
Why?

2 a⃗×b⃗2\,\vec a \times \vec b becomes 2 c⃗×d⃗2\,\vec c \times \vec d⁠, and −a⃗×c⃗-\vec a \times \vec c becomes −4 b⃗×d⃗-4\,\vec b \times \vec d⁠. Then the four terms cancel in pairs.

2 a⃗×b⃗=2 c⃗×d⃗−a⃗×c⃗=−4 b⃗×d⃗2 c⃗×d⃗−2 c⃗×d⃗=0⃗−4 b⃗×d⃗+4 b⃗×d⃗=0⃗\begin{aligned} 2\,\vec a \times \vec b &= 2\,\vec c \times \vec d \\ -\vec a \times \vec c &= -4\,\vec b \times \vec d \\[4pt] &2\,\vec c \times \vec d - 2\,\vec c \times \vec d = \vec 0 \\ &-4\,\vec b \times \vec d + 4\,\vec b \times \vec d = \vec 0 \end{aligned}

As a⃗≠2d⃗\vec a \ne 2\vec d and c⃗≠2b⃗\vec c \ne 2\vec b⁠, both vectors are non-zero and their cross product is 0⃗\vec 0⁠.

Why?

The conditions a⃗≠2d⃗\vec a \ne 2\vec d and c⃗≠2b⃗\vec c \ne 2\vec b make a⃗−2d⃗\vec a - 2\vec d and 2b⃗−c⃗2\vec b - \vec c non-zero. Without this, the cross product would be 0⃗\vec 0 simply because one vector is 0⃗\vec 0⁠, and that says nothing about direction.

∴ (a⃗−2d⃗)\therefore\ (\vec a - 2\vec d) is parallel to (2b⃗−c⃗)(2\vec b - \vec c)⁠. Hence proved.

(b)

Let ABCD be the parallelogram with

AB→=DC→=2i^−4j^−5k^\overrightarrow{AB} = \overrightarrow{DC} = 2\hat i - 4\hat j - 5\hat k BC→=AD→=2i^+2j^+3k^\overrightarrow{BC} = \overrightarrow{AD} = 2\hat i + 2\hat j + 3\hat k

The diagonals are

d⃗1=AC→=AB→+BC→=(2+2)i^+(−4+2)j^+(−5+3)k^=4i^−2j^−2k^\begin{aligned} \vec d_1 = \overrightarrow{AC} &= \overrightarrow{AB} + \overrightarrow{BC} \\ &= (2 + 2)\hat i + (-4 + 2)\hat j \\ &\quad + (-5 + 3)\hat k \\ &= 4\hat i - 2\hat j - 2\hat k \end{aligned}
Why?

Triangle law: from A to B, then B to C, is the same as A straight to C. Add the matching components.

d⃗2=BD→=AD→−AB→=(2−2)i^+(2+4)j^+(3+5)k^=6j^+8k^\begin{aligned} \vec d_2 = \overrightarrow{BD} &= \overrightarrow{AD} - \overrightarrow{AB} \\ &= (2 - 2)\hat i + (2 + 4)\hat j \\ &\quad + (3 + 5)\hat k \\ &= 6\hat j + 8\hat k \end{aligned}
Why?

From B back to A, then A to D: BD→=BA→+AD→\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD}⁠, and BA→=−AB→\overrightarrow{BA} = -\overrightarrow{AB}⁠.

Watch the signs when subtracting: 2−(−4)=62 - (-4) = 6 and 3−(−5)=83 - (-5) = 8⁠.

∣AC→∣=16+4+4=26|\overrightarrow{AC}| = \sqrt{16 + 4 + 4} = 2\sqrt6 ∣BD→∣=36+64=10|\overrightarrow{BD}| = \sqrt{36 + 64} = 10

So the required unit vectors are

d^1=d⃗1∣d⃗1∣=4i^−2j^−2k^26\hat d_1 = \frac{\vec d_1}{|\vec d_1|} = \frac{4\hat i - 2\hat j - 2\hat k}{2\sqrt6} d^1=26i^−16j^−16k^\hat d_1 = \frac{2}{\sqrt6}\hat i - \frac{1}{\sqrt6}\hat j - \frac{1}{\sqrt6}\hat k d^2=6j^+8k^10\hat d_2 = \frac{6\hat j + 8\hat k}{10} d^2=35j^+45k^\hat d_2 = \frac35\hat j + \frac45\hat k
Why?

A unit vector has length 1. Dividing a vector by its own length keeps its direction and makes the length 1:

d^=d⃗∣d⃗∣\hat d = \frac{\vec d}{|\vec d|}

Lengths: 24=4×6=26\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt6 and 100=10\sqrt{100} = 10⁠. Then divide each component.

∣d⃗1∣=(4)2+(−2)2+(−2)2=16+4+4=24=4×6=26∣d⃗2∣=(0)2+(6)2+(8)2=0+36+64=100=10\begin{aligned} |\vec d_1| &= \sqrt{(4)^2 + (-2)^2 + (-2)^2} \\ &= \sqrt{16 + 4 + 4} = \sqrt{24} \\ &= \sqrt{4 \times 6} = 2\sqrt6 \\[4pt] |\vec d_2| &= \sqrt{(0)^2 + (6)^2 + (8)^2} \\ &= \sqrt{0 + 36 + 64} \\ &= \sqrt{100} = 10 \end{aligned}

Each component of d⃗1\vec d_1 over 262\sqrt6⁠:

426=26−226=−16−226=−16\begin{aligned} \frac{4}{2\sqrt6} &= \frac{2}{\sqrt6} \\ \frac{-2}{2\sqrt6} &= -\frac{1}{\sqrt6} \\ \frac{-2}{2\sqrt6} &= -\frac{1}{\sqrt6} \end{aligned}

Each component of d⃗2\vec d_2 over 10:

610=6÷210÷2=35810=8÷210÷2=45\begin{aligned} \frac{6}{10} &= \frac{6 \div 2}{10 \div 2} = \frac35 \\ \frac{8}{10} &= \frac{8 \div 2}{10 \div 2} = \frac45 \end{aligned}

Area of parallelogram =12∣d⃗1×d⃗2∣= \frac12|\vec d_1 \times \vec d_2|⁠.

Why?

With adjacent sides a⃗\vec a and b⃗\vec b⁠, the diagonals are a⃗+b⃗\vec a + \vec b and b⃗−a⃗\vec b - \vec a⁠. Since a⃗×a⃗=b⃗×b⃗=0⃗\vec a \times \vec a = \vec b \times \vec b = \vec 0⁠,

(a⃗+b⃗)×(b⃗−a⃗)=a⃗×b⃗−b⃗×a⃗=2 a⃗×b⃗\begin{aligned} &(\vec a + \vec b) \times (\vec b - \vec a) \\ &= \vec a \times \vec b - \vec b \times \vec a \\ &= 2\,\vec a \times \vec b \end{aligned}

So half of ∣d⃗1×d⃗2∣|\vec d_1 \times \vec d_2| equals ∣a⃗×b⃗∣|\vec a \times \vec b|⁠, the area.

d⃗1×d⃗2=∣i^j^k^4−2−2068∣\vec d_1 \times \vec d_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & -2 & -2 \\ 0 & 6 & 8 \end{vmatrix} =i^(−16+12)−j^(32−0)+k^(24−0)=−4i^−32j^+24k^\begin{aligned} &= \hat i(-16 + 12) - \hat j(32 - 0) \\ &\quad + \hat k(24 - 0) \\ &= -4\hat i - 32\hat j + 24\hat k \end{aligned}
Why?

Expand along the first row, with signs ++⁠, −-⁠, ++⁠. Each bracket is a 2 by 2 determinant:

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^: −16−(−12)=−16+12=−4j^: 32−0=32k^: 24−0=24\begin{aligned} \hat i&:\ -16 - (-12) = -16 + 12 \\ &\quad = -4 \\ \hat j&:\ 32 - 0 = 32 \\ \hat k&:\ 24 - 0 = 24 \end{aligned}

Watch out for the minus sign in front of the j^\hat j term: −j^(32)=−32j^-\hat j(32) = -32\hat j⁠.

Area=1216+1024+576=121616=12×4101\begin{aligned} \text{Area} &= \frac12\sqrt{16 + 1024 + 576} \\ &= \frac12\sqrt{1616} = \frac12 \times 4\sqrt{101} \end{aligned}
Why?

Square each component: (−4)2=16(-4)^2 = 16⁠, (−32)2=1024(-32)^2 = 1024⁠, 242=57624^2 = 576⁠.

16+1024=10401040+576=1616\begin{aligned} 16 + 1024 &= 1040 \\ 1040 + 576 &= 1616 \end{aligned}

Then 1616=16×1011616 = 16 \times 101⁠, so 1616=4101\sqrt{1616} = 4\sqrt{101}⁠.

1616=16×101=16×101=410112×4101=2101\begin{aligned} \sqrt{1616} &= \sqrt{16 \times 101} \\ &= \sqrt{16} \times \sqrt{101} = 4\sqrt{101} \\[4pt] \frac12 \times 4\sqrt{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: Cross product: parallel vectors, area.

Question type: (a) Dot products with letters; (b) A unit vector, or a vector of a given length + Cross product, and the area of a parallelogram (combined)

How to spot it: click to see

(a) Type 7. Dot products with letters: expand like brackets

Spot it: no i^,j^,k^\hat i, \hat j, \hat k at all, just a⃗\vec a⁠, b⃗\vec b⁠, their lengths and a dot product.

Here: There are no i^,j^,k^\hat i, \hat j, \hat k⁠, only letters: expand (a⃗−2d⃗)×(2b⃗−c⃗)(\vec a - 2\vec d) \times (2\vec b - \vec c) like brackets, keeping the order inside each cross product, then put in the two given conditions to get 0⃗\vec 0⁠.

(b) Type 1. A unit vector, or a vector of a given length

Spot it: "unit vector in the direction of", or "a vector of magnitude 8 in the direction of".

Here: First, the unit vectors: the diagonals are 4i^−2j^−2k^4\hat i - 2\hat j - 2\hat k and 6j^+8k^6\hat j + 8\hat k⁠, so divide each by its length, 262\sqrt6 and 10.

(b) 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: Then the area: find d⃗1×d⃗2\vec d_1 \times \vec d_2 by the determinant and take half its length, because these are diagonals, not sides.

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