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CBSE Sample Paper 2024-25 · Section C · 3 marks

Q28: Angle between l1⃗\vec{l_1} and l2⃗\vec{l_2} and the projection, or the line perpendicular to two given lines

Sample 2024-25Ch 10 · Vector AlgebraCh 11 · Three Dimensional Geometry
The question

(a) An ant is moving along the vector l1⃗=i^−2j^+3k^\vec{l_1} = \hat i - 2\hat j + 3\hat k⁠. Few sugar crystals are kept along the vector l2⃗=3i^−2j^+k^\vec{l_2} = 3\hat i - 2\hat j + \hat k which is inclined at an angle θ\theta with the vector l1⃗\vec{l_1}⁠. Then find the angle θ\theta⁠. Also find the scalar projection of l1⃗\vec{l_1} on l2⃗\vec{l_2}⁠.

OR

(b) Find the vector and the cartesian equation of the line that passes through (−1,2,7)(-1, 2, 7) and is perpendicular to the lines

r⃗=2i^+j^−3k^+λ(i^+2j^+5k^)\begin{aligned} \vec r &= 2\hat i + \hat j - 3\hat k \\ &\quad + \lambda(\hat i + 2\hat j + 5\hat k) \end{aligned}

and

r⃗=3i^+3j^−7k^+μ(3i^−2j^+5k^).\begin{aligned} \vec r &= 3\hat i + 3\hat j - 7\hat k \\ &\quad + \mu(3\hat i - 2\hat j + 5\hat k). \end{aligned}
Exam answer

(a)

θ=cos⁡−1(l1⃗⋅l2⃗∣l1⃗∣∣l2⃗∣)\theta = \cos^{-1}\left(\frac{\vec{l_1}\cdot\vec{l_2}}{\lvert\vec{l_1}\rvert\lvert\vec{l_2}\rvert}\right)
Why?

This comes from the definition of the dot product:

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

Divide by the two lengths, then take cos⁡−1\cos^{-1}⁠. The angle between two vectors lies in [0,π][0, \pi]⁠, which is exactly the range of cos⁡−1\cos^{-1}⁠.

l1⃗⋅l2⃗=3+4+3=10\vec{l_1}\cdot\vec{l_2} = 3 + 4 + 3 = 10
Why?

Multiply matching components and add:

(1)(3)+(−2)(−2)+(3)(1)=3+4+3=7+3=10\begin{aligned} &(1)(3) + (-2)(-2) + (3)(1) \\ &= 3 + 4 + 3 \\ &= 7 + 3 = 10 \end{aligned}

Watch the sign: (−2)(−2)=+4(-2)(-2) = +4⁠.

∣l1⃗∣=1+4+9=14\lvert\vec{l_1}\rvert = \sqrt{1 + 4 + 9} = \sqrt{14} ∣l2⃗∣=9+4+1=14\lvert\vec{l_2}\rvert = \sqrt{9 + 4 + 1} = \sqrt{14}
Why?

The length of ai^+bj^+ck^a\hat i + b\hat j + c\hat k is a2+b2+c2\sqrt{a^2 + b^2 + c^2}⁠. Squares are never negative: (−2)2=4(-2)^2 = 4⁠.

∣l1⃗∣2=12+(−2)2+32=1+4+9=14∣l2⃗∣2=32+(−2)2+12=9+4+1=14\begin{aligned} \lvert\vec{l_1}\rvert^2 &= 1^2 + (-2)^2 + 3^2 \\ &= 1 + 4 + 9 = 14 \\[4pt] \lvert\vec{l_2}\rvert^2 &= 3^2 + (-2)^2 + 1^2 \\ &= 9 + 4 + 1 = 14 \end{aligned}
θ=cos⁡−1(101414)=cos⁡−1(1014)\begin{aligned} \theta &= \cos^{-1}\left(\frac{10}{\sqrt{14}\sqrt{14}}\right) \\ &= \cos^{-1}\left(\frac{10}{14}\right) \end{aligned}
Why?

14⋅14=14\sqrt{14}\cdot\sqrt{14} = 14⁠, and 1014=57\frac{10}{14} = \frac57 after dividing top and bottom by 2.

1014=10÷214÷2=57\frac{10}{14} = \frac{10 \div 2}{14 \div 2} = \frac57

∴ θ=cos⁡−1(57)\therefore\ \theta = \cos^{-1}\left(\frac57\right)

Scalar projection of l1⃗\vec{l_1} on l2⃗\vec{l_2}

=l1⃗⋅l2⃗∣l2⃗∣=3+4+39+4+1=1014\begin{aligned} &= \frac{\vec{l_1}\cdot\vec{l_2}}{\lvert\vec{l_2}\rvert} \\ &= \frac{3 + 4 + 3}{\sqrt{9 + 4 + 1}} \\ &= \frac{10}{\sqrt{14}} \end{aligned}
Why?

The scalar projection of a⃗\vec a on b⃗\vec b is a⃗⋅b^\vec a\cdot\hat b⁠, the dot product with the unit vector along b⃗\vec b⁠:

a⃗⋅b^=a⃗⋅b⃗∣b⃗∣\vec a\cdot\hat b = \frac{\vec a\cdot\vec b}{\lvert\vec b\rvert}

Watch out: projecting on l2⃗\vec{l_2} means dividing by ∣l2⃗∣\lvert\vec{l_2}\rvert⁠, not ∣l1⃗∣\lvert\vec{l_1}\rvert⁠. (Here both are 14\sqrt{14}⁠.)

The scalar projection of l1⃗\vec{l_1} on l2⃗\vec{l_2} is 1014\dfrac{10}{\sqrt{14}}⁠.

(b)

Here b1⃗=i^+2j^+5k^\vec{b_1} = \hat i + 2\hat j + 5\hat k and b2⃗=3i^−2j^+5k^\vec{b_2} = 3\hat i - 2\hat j + 5\hat k⁠.

Why?

b1⃗\vec{b_1} and b2⃗\vec{b_2} are the vectors multiplying λ\lambda and μ\mu⁠: they give the directions of the two lines.

The cross product b1⃗×b2⃗\vec{b_1} \times \vec{b_2} is perpendicular to both b1⃗\vec{b_1} and b2⃗\vec{b_2}⁠, so it is a direction for the required line.

The required line is perpendicular to both lines, so it is parallel to

b⃗=b1⃗×b2⃗=∣i^j^k^1253−25∣=i^(10+10)−j^(5−15)+k^(−2−6)=20i^+10j^−8k^=2(10i^+5j^−4k^)\begin{aligned} \vec b &= \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 5 \\ 3 & -2 & 5 \end{vmatrix} \\ &= \hat i(10 + 10) - \hat j(5 - 15) \\ &\quad + \hat k(-2 - 6) \\ &= 20\hat i + 10\hat j - 8\hat k \\ &= 2(10\hat i + 5\hat j - 4\hat k) \end{aligned}
Why?

Expand along the first row with the signs ++⁠, −-⁠, ++⁠:

i^:(2)(5)−(5)(−2)=20j^:(1)(5)−(5)(3)=−10k^:(1)(−2)−(2)(3)=−8\begin{aligned} \hat i&: (2)(5) - (5)(-2) = 20 \\ \hat j&: (1)(5) - (5)(3) = -10 \\ \hat k&: (1)(-2) - (2)(3) = -8 \end{aligned}

Each one in steps (cover the row of unit vectors and that column):

i^:∣25−25∣=(2)(5)−(5)(−2)=10−(−10)=10+10=20j^:∣1535∣=(1)(5)−(5)(3)=5−15=−10k^:∣123−2∣=(1)(−2)−(2)(3)=−2−6=−8\begin{aligned} \hat i&: \begin{vmatrix} 2 & 5 \\ -2 & 5 \end{vmatrix} \\ &= (2)(5) - (5)(-2) \\ &= 10 - (-10) = 10 + 10 = 20 \\[4pt] \hat j&: \begin{vmatrix} 1 & 5 \\ 3 & 5 \end{vmatrix} \\ &= (1)(5) - (5)(3) \\ &= 5 - 15 = -10 \\[4pt] \hat k&: \begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix} \\ &= (1)(-2) - (2)(3) \\ &= -2 - 6 = -8 \end{aligned}

The j^\hat j term has a minus sign in front, so it becomes +10j^+10\hat j⁠.

−j^(−10)=+10j^20i^+10j^−8k^=2(10i^)+2(5j^)−2(4k^)=2(10i^+5j^−4k^)\begin{aligned} -\hat j(-10) &= +10\hat j \\[4pt] 20\hat i + 10\hat j - 8\hat k &= 2(10\hat i) + 2(5\hat j) \\ &\quad - 2(4\hat k) \\ &= 2(10\hat i + 5\hat j - 4\hat k) \end{aligned}

Any non-zero multiple gives the same direction, so the factor 2 can be dropped.

∴\therefore Equation of the line in vector form is

r⃗=−i^+2j^+7k^+t(10i^+5j^−4k^)\begin{aligned} \vec r &= -\hat i + 2\hat j + 7\hat k \\ &\quad + t(10\hat i + 5\hat j - 4\hat k) \end{aligned}
Why?

A line through the point with position vector a⃗\vec a⁠, parallel to b⃗\vec b⁠, is r⃗=a⃗+tb⃗\vec r = \vec a + t\vec b⁠.

The point (−1,2,7)(-1, 2, 7) has position vector −i^+2j^+7k^-\hat i + 2\hat j + 7\hat k⁠.

where tt is a scalar.

The line passes through (−1,2,7)(-1, 2, 7) and has direction ratios 10, 5, −4-4⁠.

Why?

A line through (x1,y1,z1)(x_1, y_1, z_1) with direction ratios aa⁠, bb⁠, cc is

x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

Here x−(−1)=x+1x - (-1) = x + 1⁠. The direction ratios are the components of 10i^+5j^−4k^10\hat i + 5\hat j - 4\hat k⁠.

∴\therefore The equation of the line in cartesian form is

x+110=y−25=z−7−4\dfrac{x + 1}{10} = \dfrac{y - 2}{5} = \dfrac{z - 7}{-4}⁠.

Where this comes from

Chapter 10, Vector Algebra: Angle, projection; line perpendicular to two lines.

Question type: (a) The angle between two vectors + Projection (combined); (b) A line perpendicular to two lines

How to spot it: click to see

(a) Ch 10, type 4. The angle between two vectors

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

Here: First the angle: l1⃗⋅l2⃗=10\vec{l_1}\cdot\vec{l_2} = 10 and both lengths are 14\sqrt{14}⁠, so cos⁡θ=57\cos\theta = \frac57⁠.

(a) Ch 10, type 6. Projection: the shadow of one vector on another

Spot it: "find the projection of the vector a⃗\vec a on the vector b⃗\vec b".

Here: Then the projection of l1⃗\vec{l_1} on l2⃗\vec{l_2}⁠: reuse the dot product 10 and divide by ∣l2⃗∣=14|\vec{l_2}| = \sqrt{14}⁠.

(b) Ch 11, type 9. A line perpendicular to two lines: use the cross product

Spot it: "Find the line through the point … perpendicular to the two lines …". The CBSE sample papers for 2024-25, 2025-26 and 2026-27 all ask it.

Here: The line must be perpendicular to both given lines, so its direction is b1⃗×b2⃗=2(10i^+5j^−4k^)\vec{b_1} \times \vec{b_2} = 2(10\hat i + 5\hat j - 4\hat k)⁠. Write it through the point (−1,2,7)(-1, 2, 7)⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 11, Three Dimensional Geometry.

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