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CBSE Board Examination 2023 · Section A · 1 mark

Q8: p for which 2i^+pj^+k^2\hat i + p\hat j + \hat k and −4i^−6j^+26k^-4\hat i - 6\hat j + 26\hat k are perpendicular

Board 2023Ch 10 · Vector Algebra
The question

The value of p for which the vectors 2i^+pj^+k^2\hat i + p\hat j + \hat k and −4i^−6j^+26k^-4\hat i - 6\hat j + 26\hat k are perpendicular to each other, is :

(a) 33

(b) −3-3

(c) −173-\frac{17}{3}

(d) 173\frac{17}{3}

Exam answer

Perpendicular vectors have scalar product 0:

Why?

The scalar product is a⃗⋅b⃗=∣a⃗∣ ∣b⃗∣cos⁡θ\vec a \cdot \vec b = |\vec a|\,|\vec b|\cos\theta⁠. For perpendicular vectors θ=90∘\theta = 90^\circ and cos⁡90∘=0\cos 90^\circ = 0⁠, so the product is 0.

In components, multiply matching coefficients and add:

a⃗⋅b⃗=a1b1+a2b2+a3b3\begin{aligned} \vec a \cdot \vec b &= a_1b_1 + a_2b_2 \\ &\quad + a_3b_3 \end{aligned}
2(−4)+p(−6)+1(26)=0⇒ −8−6p+26=0⇒ 18−6p=0⇒ 6p=18⇒ p=3\begin{aligned} &2(-4) + p(-6) + 1(26) = 0 \\ \Rightarrow\ &-8 - 6p + 26 = 0 \\ \Rightarrow\ &18 - 6p = 0 \\ \Rightarrow\ &6p = 18 \\ \Rightarrow\ &p = 3 \end{aligned}
Why?

The coefficients are 2,p,12, p, 1 for the first vector and −4,−6,26-4, -6, 26 for the second. The three products:

2(−4)=−8p(−6)=−6p1(26)=26\begin{aligned} 2(-4) &= -8 \\ p(-6) &= -6p \\ 1(26) &= 26 \end{aligned}

Then −8+26=18-8 + 26 = 18⁠. Solve one move at a time:

18−6p=018=6pp=186=3\begin{aligned} 18 - 6p &= 0 \\ 18 &= 6p \\ p &= \frac{18}{6} = 3 \end{aligned}

Check: with p=3p = 3⁠, −8−18+26=0-8 - 18 + 26 = 0⁠.

2(−4)+3(−6)+1(26)=−8−18+26=−26+26=0\begin{aligned} &2(-4) + 3(-6) + 1(26) \\ &= -8 - 18 + 26 \\ &= -26 + 26 = 0 \end{aligned}

Watch out: a sign slip in p(−6)p(-6) gives option (b), −3-3⁠.

Answer: (a) 33

Where this comes from

Chapter 10, Vector Algebra: Perpendicular vectors: scalar product.

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 pp sits in the j^\hat j part of the first vector. Set the dot product −8−6p+26-8 - 6p + 26 equal to 0, so p=3p = 3⁠.

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