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CBSE Sample Paper 2023-24 · Section A · 1 mark

Q16: λ\lambda for which 2i^−j^+2k^2\hat i - \hat j + 2\hat k and 3i^+λj^+k^3\hat i + \lambda\hat j + \hat k are perpendicular

Sample 2023-24Ch 10 · Vector Algebra
The question

The value of λ\lambda for which two vectors 2i^−j^+2k^2\hat i - \hat j + 2\hat k and 3i^+λj^+k^3\hat i + \lambda\hat j + \hat k are perpendicular is

(a) 22

(b) 44

(c) 66

(d) 88

Exam answer

Perpendicular vectors have dot product zero.

Why?

a⃗⋅b⃗=∣a⃗∣ ∣b⃗∣cos⁡θ\vec a \cdot \vec b = |\vec a|\,|\vec b|\cos\theta⁠. For non-zero vectors this is 0 exactly when cos⁡θ=0\cos\theta = 0⁠, that is, when θ=π2\theta = \frac{\pi}{2}⁠.

(2)(3)+(−1)(λ)+(2)(1)=0⇒ 8−λ=0\begin{aligned} &(2)(3) + (-1)(\lambda) + (2)(1) = 0 \\ \Rightarrow\ &8 - \lambda = 0 \end{aligned}
Why?

Multiply the matching components and add:

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a \cdot \vec b = a_1b_1 + a_2b_2 + a_3b_3

Here that is 6−λ+2=8−λ6 - \lambda + 2 = 8 - \lambda⁠.

(2)(3)=6(−1)(λ)=−λ(2)(1)=26−λ+2=8−λ(6+2=8)\begin{aligned} (2)(3) &= 6 \\ (-1)(\lambda) &= -\lambda \\ (2)(1) &= 2 \\[4pt] 6 - \lambda + 2 &= 8 - \lambda \quad (6 + 2 = 8) \end{aligned}

∴ λ=8\therefore\ \lambda = 8

Why?

Add λ\lambda to both sides of 8−λ=08 - \lambda = 0⁠:

8−λ+λ=0+λ8=λ\begin{aligned} 8 - \lambda + \lambda &= 0 + \lambda \\ 8 &= \lambda \end{aligned}

Check with λ=8\lambda = 8⁠: 6−8+2=06 - 8 + 2 = 0⁠.

Watch out: the j^\hat j component of the first vector is −1-1⁠, so the middle term is −λ-\lambda⁠, not +λ+\lambda⁠.

Answer: (d) 88

Where this comes from

Chapter 10, Vector Algebra: Perpendicular vectors: dot product zero.

Question type: Perpendicular vectors

How to spot it: click to see

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

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

Here: The unknown λ\lambda sits in the j^\hat j part of 3i^+λj^+k^3\hat i + \lambda\hat j + \hat k⁠. Set the dot product 6−λ+26 - \lambda + 2 equal to 0, so λ=8\lambda = 8⁠.

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