CBSE Sample Paper 2024-25 · Section B · 2 marks

Q24: $\lambda$ for $\vec b + \lambda\vec c \perp \vec a$ ⁠, or the angles $\overrightarrow{BA}$ makes with the axes

Sample 2024-25Ch 10 · Vector Algebra

The question

(a) If vectors $\vec a = 2\hat i + 2\hat j + 3\hat k$ ⁠, $\vec b = -\hat i + 2\hat j + \hat k$ and $\vec c = 3\hat i + \hat j$ are such that $\vec b + \lambda\vec c$ is perpendicular to $\vec a$ ⁠, then find the value of $\lambda$ ⁠.

(b) A person standing at $O(0, 0, 0)$ is watching an aeroplane which is at the coordinate point $A(4, 0, 3)$ ⁠. At the same time he saw a bird at the coordinate point $B(0, 0, 1)$ ⁠. Find the angles which $\overrightarrow{BA}$ makes with the x,y and z axes.

Exam answer

(a)

\begin{aligned} &\vec b + \lambda\vec c \\ &= (-\hat i + 2\hat j + \hat k) + \lambda(3\hat i + \hat j) \\ &= (-1 + 3\lambda)\hat i + (2 + \lambda)\hat j + \hat k \end{aligned}

Why?

Multiply each component of $\vec c$ by $\lambda$ ⁠, then add like components. $\vec c$ has no $\hat k$ part, so the $\hat k$ component stays 1:

\vec c = 3\hat i + \hat j + 0\hat k

\begin{aligned} \lambda\vec c &= 3\lambda\hat i + \lambda\hat j + 0\hat k \\[4pt] \hat i&: -1 + 3\lambda \\ \hat j&: 2 + \lambda \\ \hat k&: 1 + 0 = 1 \end{aligned}

$(\vec b + \lambda\vec c)\cdot\vec a = 0$ (perpendicular)

Why?

Two non-zero vectors are perpendicular exactly when their dot product is 0. The dot product multiplies matching components and adds:

\vec p\cdot\vec q = p_1q_1 + p_2q_2 + p_3q_3

Here the components of $\vec a$ are 2, 2 and 3.

\begin{aligned} &\Rightarrow 2(-1 + 3\lambda) + 2(2 + \lambda) \\ &\qquad + 3 = 0 \\ &\Rightarrow -2 + 6\lambda + 4 + 2\lambda + 3 = 0 \\ &\Rightarrow 8\lambda + 5 = 0 \end{aligned}

Why?

Open each bracket first:

\begin{aligned} 2(-1 + 3\lambda) &= -2 + 6\lambda \\ 2(2 + \lambda) &= 4 + 2\lambda \\ 3(1) &= 3 \end{aligned}

Collect like terms: $6\lambda + 2\lambda = 8\lambda$ ⁠, and $-2 + 4 + 3 = 5$ ⁠.

\begin{aligned} -2 + 4 &= 2 \\ 2 + 3 &= 5 \end{aligned}

Then $8\lambda = -5$ ⁠, so $\lambda = -\frac58$ ⁠.

\begin{aligned} 8\lambda + 5 &= 0 \\ 8\lambda &= -5 \quad (-5 \text{ both sides}) \\ \lambda &= -\frac58 \quad (\div\,8) \end{aligned}

$\therefore\ \lambda = -\dfrac58$

(b)

\begin{aligned} \overrightarrow{BA} &= \overrightarrow{OA} - \overrightarrow{OB} \\ &= (4\hat i + 3\hat k) - \hat k \\ &= 4\hat i + 2\hat k \end{aligned}

Why?

A vector from B to A is the position vector of its head A minus the position vector of its tail B. With O at the origin,

\begin{aligned} \overrightarrow{OA} &= 4\hat i + 0\hat j + 3\hat k \\ \overrightarrow{OB} &= 0\hat i + 0\hat j + \hat k \end{aligned}

Subtract component by component:

\begin{aligned} \hat i&: 4 - 0 = 4 \\ \hat j&: 0 - 0 = 0 \\ \hat k&: 3 - 1 = 2 \end{aligned}

\lvert\overrightarrow{BA}\rvert = \sqrt{16 + 4} = 2\sqrt5

Why?

The magnitude is $\sqrt{4^2 + 0^2 + 2^2} = \sqrt{20}$ ⁠, and $\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt5$ ⁠.

\begin{aligned} \widehat{BA} &= \frac{4}{2\sqrt5}\hat i + \frac{2}{2\sqrt5}\hat k \\ &= \frac{2}{\sqrt5}\hat i + 0\hat j + \frac{1}{\sqrt5}\hat k \end{aligned}

Why?

The unit vector along a vector is the vector divided by its magnitude.

\begin{aligned} \widehat{BA} &= \frac{4\hat i + 2\hat k}{2\sqrt5} \\[4pt] \frac{4}{2\sqrt5} &= \frac{4 \div 2}{2\sqrt5 \div 2} = \frac{2}{\sqrt5} \\[4pt] \frac{2}{2\sqrt5} &= \frac{2 \div 2}{2\sqrt5 \div 2} = \frac{1}{\sqrt5} \end{aligned}

The components of this unit vector are the direction cosines $\cos\alpha$ ⁠, $\cos\beta$ ⁠, $\cos\gamma$ ⁠, where $\alpha$ ⁠, $\beta$ ⁠, $\gamma$ are the angles the vector makes with the $x$ ⁠, $y$ and $z$ axes.

If $\overrightarrow{BA}$ makes angles $\alpha$ ⁠, $\beta$ ⁠, $\gamma$ with the $x$ ⁠, $y$ and $z$ axes, then

\begin{aligned} &\cos\alpha = \frac{2}{\sqrt5},\ \cos\beta = 0, \\ &\cos\gamma = \frac{1}{\sqrt5} \end{aligned}

Why?

Each of these angles lies between 0 and $\pi$ ⁠, so each angle is the $\cos^{-1}$ of its direction cosine.

$\cos\beta = 0$ gives $\beta = \frac{\pi}{2}$ ⁠. This fits: $\overrightarrow{BA}$ has no $\hat j$ part, so it is perpendicular to the $y$ -axis.

Check: the squares of the direction cosines add up to 1.

\begin{aligned} &\cos^2\alpha + \cos^2\beta + \cos^2\gamma \\ &= \left(\frac{2}{\sqrt5}\right)^2 + 0^2 + \left(\frac{1}{\sqrt5}\right)^2 \\ &= \frac45 + 0 + \frac15 = 1 \end{aligned}

So, the angles made by $\overrightarrow{BA}$ with the $x$ ⁠, $y$ and $z$ axes are $\cos^{-1}\left(\dfrac{2}{\sqrt5}\right)$ ⁠, $\dfrac{\pi}{2}$ and $\cos^{-1}\left(\dfrac{1}{\sqrt5}\right)$ respectively.

Where this comes from

Chapter 10, Vector Algebra: Perpendicular vectors; angles with the axes.

Question type: (a) Perpendicular vectors; (b) The vector from A to B, and its direction cosines

How to spot it: click to see

(a) Type 5. Perpendicular vectors: set the dot product to 0

Spot it: "is perpendicular to", with an unknown such as $\lambda$ to find.

Here: $\vec b + \lambda\vec c = (-1 + 3\lambda)\hat i + (2 + \lambda)\hat j + \hat k$ must be perpendicular to $\vec a$ ⁠. Set its dot product with $\vec a$ to 0: $8\lambda + 5 = 0$ ⁠.

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

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

Here: From the two points, $\overrightarrow{BA} = 4\hat i + 2\hat k$ ⁠, of length $2\sqrt5$ ⁠. Its direction cosines $\frac{2}{\sqrt5}, 0, \frac{1}{\sqrt5}$ are the cosines of the angles with the axes.

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Vector Algebra in past papers: question 31 of 48 · see them all

Q24: λ\lambdaλ for b⃗+λc⃗⊥a⃗\vec b + \lambda\vec c \perp \vec ab+λc⊥a⁠, or the angles BA→\overrightarrow{BA}BA makes with the axes

Q24: $\lambda$ for $\vec b + \lambda\vec c \perp \vec a$ ⁠, or the angles $\overrightarrow{BA}$ makes with the axes