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CBSE Board Examination 2025 · Section B · 2 marks

Q25: Angle between kite strings 3i^+j^+2k^3\hat i + \hat j + 2\hat k and 2i^−2j^+4k^2\hat i - 2\hat j + 4\hat k⁠, or a vector of magnitude 21 opposite to AB→\overrightarrow{AB}

Board 2025Ch 10 · Vector Algebra
The question

(a) Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors a⃗=3i^+j^+2k^\vec a = 3\hat i + \hat j + 2\hat k and b⃗=2i^−2j^+4k^\vec b = 2\hat i - 2\hat j + 4\hat k⁠. Determine the angle formed between the kite strings. Assume there is no slack in the strings.

OR

(b) Find a vector of magnitude 21 units in the direction opposite to that of AB→\overrightarrow{AB} where A and B are the points A(2, 1, 3) and B(8, −1-1⁠, 0) respectively.

Exam answer

(a)

Let θ\theta be the angle between the kite strings. Then

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}
Why?

The strings lie along a⃗\vec a and b⃗\vec b⁠, so the angle between the strings is the angle between the vectors. It comes from the definition of the dot product:

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta

Divide both sides by ∣a⃗∣∣b⃗∣|\vec a||\vec b|⁠.

a⃗⋅b⃗=3(2)+1(−2)+2(4)=6−2+8=12\begin{aligned} \vec a\cdot\vec b &= 3(2) + 1(-2) + 2(4) \\ &= 6 - 2 + 8 = 12 \end{aligned}
Why?

For a dot product, multiply matching components and add:

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3
∣a⃗∣=9+1+4=14|\vec a| = \sqrt{9 + 1 + 4} = \sqrt{14} ∣b⃗∣=4+4+16=24|\vec b| = \sqrt{4 + 4 + 16} = \sqrt{24}
Why?

The magnitude is the square root of the sum of the squares of the components: 32,12,223^2, 1^2, 2^2 for a⃗\vec a⁠, and 22,(−2)2,422^2, (-2)^2, 4^2 for b⃗\vec b⁠.

∣a⃗∣=(3)2+(1)2+(2)2=9+1+4=14∣b⃗∣=(2)2+(−2)2+(4)2=4+4+16=24\begin{aligned} |\vec a| &= \sqrt{(3)^2 + (1)^2 + (2)^2} \\ &= \sqrt{9 + 1 + 4} = \sqrt{14} \\[4pt] |\vec b| &= \sqrt{(2)^2 + (-2)^2 + (4)^2} \\ &= \sqrt{4 + 4 + 16} = \sqrt{24} \end{aligned}
cos⁡θ=121424=12336=12421=321\begin{aligned} \cos\theta &= \frac{12}{\sqrt{14}\sqrt{24}} = \frac{12}{\sqrt{336}} \\ &= \frac{12}{4\sqrt{21}} = \frac{3}{\sqrt{21}} \end{aligned}
Why?

14×24=336=16×2114\times24 = 336 = 16\times21⁠, so 336=421\sqrt{336} = 4\sqrt{21}⁠. Then cancel 44 from 1212⁠.

1424=14×24=336336=16×21=16 21=42112421=12÷421=321\begin{aligned} \sqrt{14}\sqrt{24} &= \sqrt{14 \times 24} = \sqrt{336} \\ \sqrt{336} &= \sqrt{16 \times 21} \\ &= \sqrt{16}\,\sqrt{21} = 4\sqrt{21} \\[4pt] \frac{12}{4\sqrt{21}} &= \frac{12 \div 4}{\sqrt{21}} = \frac{3}{\sqrt{21}} \end{aligned}

cos⁡θ\cos\theta is positive, so the angle is acute.

∴\therefore The angle is θ=cos⁡−1(321)\theta = \cos^{-1}\left(\frac{3}{\sqrt{21}}\right)⁠.

(b)

AB→=(8−2)i^+(−1−1)j^+(0−3)k^=6i^−2j^−3k^\begin{aligned} \overrightarrow{AB} &= (8 - 2)\hat i + (-1 - 1)\hat j \\ &\quad + (0 - 3)\hat k \\ &= 6\hat i - 2\hat j - 3\hat k \end{aligned}
Why?

The vector from AA to BB is the position vector of BB minus the position vector of AA⁠: "end minus start", component by component.

The direction opposite to AB→\overrightarrow{AB} is that of

BA→=−6i^+2j^+3k^\overrightarrow{BA} = -6\hat i + 2\hat j + 3\hat k
Why?

Reversing a vector changes the sign of every component: BA→=−AB→\overrightarrow{BA} = -\overrightarrow{AB}⁠.

∣BA→∣=36+4+9=7|\overrightarrow{BA}| = \sqrt{36 + 4 + 9} = 7
Why?

Square each component of BA→\overrightarrow{BA}⁠, add, then take the square root:

∣BA→∣=(−6)2+(2)2+(3)2=36+4+9=49=7\begin{aligned} |\overrightarrow{BA}| &= \sqrt{(-6)^2 + (2)^2 + (3)^2} \\ &= \sqrt{36 + 4 + 9} \\ &= \sqrt{49} = 7 \end{aligned}

Required vector

=21(−6i^+2j^+3k^7)=3(−6i^+2j^+3k^)\begin{aligned} &= 21\left(\frac{-6\hat i + 2\hat j + 3\hat k}{7}\right) \\ &= 3(-6\hat i + 2\hat j + 3\hat k) \end{aligned}
Why?

Dividing BA→\overrightarrow{BA} by its length 77 gives a unit vector (length 11⁠) in the required direction. Multiplying it by 2121 gives length 2121 in the same direction:

v⃗=21⋅BA→∣BA→∣\vec v = 21\cdot\frac{\overrightarrow{BA}}{|\overrightarrow{BA}|}

217=3\frac{21}{7} = 3⁠. Then multiply each component by 3:

3(−6i^)=−18i^3(2j^)=6j^3(3k^)=9k^\begin{aligned} 3(-6\hat i) &= -18\hat i \\ 3(2\hat j) &= 6\hat j \\ 3(3\hat k) &= 9\hat k \end{aligned}

Check: its length is 3×7=213 \times 7 = 21⁠.

∴\therefore Required vector =−18i^+6j^+9k^= -18\hat i + 6\hat j + 9\hat k⁠.

Where this comes from

Chapter 10, Vector Algebra: Angle between vectors; vector of given magnitude.

Question type: (a) The angle between two vectors; (b) The vector from A to B, and its direction cosines + A unit vector, or a vector of a given length (combined)

How to spot it: click to see

(a) Type 4. The angle between two vectors

Spot it: "find the angle between the vectors".

Here: The strings are a⃗\vec a and b⃗\vec b⁠: find a⃗⋅b⃗=12\vec a\cdot\vec b = 12 and the lengths 14\sqrt{14} and 24\sqrt{24}⁠, so cos⁡θ=321\cos\theta = \frac{3}{\sqrt{21}}⁠.

(b) Type 2. The vector from A to B, and its direction cosines

Spot it: two points are given, like A(1,2,−3)A(1, 2, -3) and B(−1,−2,1)B(-1, -2, 1)⁠.

Here: First, the points A and B give AB→=6i^−2j^−3k^\overrightarrow{AB} = 6\hat i - 2\hat j - 3\hat k⁠. The opposite direction is BA→\overrightarrow{BA}⁠, with every sign changed.

(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: Then the length 21: divide BA→\overrightarrow{BA} by its length 7 and multiply by 21.

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