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CBSE Board Examination 2023 · Section A · 1 mark

Q9: (i^×j^)⋅j^+(j^×i^)⋅k^(\hat i \times \hat j) \cdot \hat j + (\hat j \times \hat i) \cdot \hat k

Board 2023Ch 10 · Vector Algebra
The question

The value of (i^×j^)⋅j^+(j^×i^)⋅k^(\hat i \times \hat j) \cdot \hat j + (\hat j \times \hat i) \cdot \hat k is :

(a) 22

(b) 00

(c) 11

(d) −1-1

Exam answer

i^×j^=k^\hat i \times \hat j = \hat k and j^×i^=−k^\hat j \times \hat i = -\hat k⁠.

Why?

The cross products of the unit vectors follow the cycle i^→j^→k^→i^\hat i \to \hat j \to \hat k \to \hat i⁠. Going with the cycle gives a plus sign: i^×j^=k^\hat i \times \hat j = \hat k⁠.

Reversing the order changes the sign, because b⃗×a⃗=−(a⃗×b⃗)\vec b \times \vec a = -(\vec a \times \vec b)⁠. So j^×i^=−k^\hat j \times \hat i = -\hat k⁠.

Also k^⋅j^=0\hat k \cdot \hat j = 0 and k^⋅k^=1\hat k \cdot \hat k = 1⁠.

Why?

k^\hat k and j^\hat j are perpendicular, so their scalar product is 0.

A unit vector dotted with itself gives its length squared: k^⋅k^=∣k^∣2=1\hat k \cdot \hat k = |\hat k|^2 = 1⁠.

(i^×j^)⋅j^+(j^×i^)⋅k^=k^⋅j^−k^⋅k^=0−1=−1\begin{aligned} &(\hat i \times \hat j) \cdot \hat j + (\hat j \times \hat i) \cdot \hat k \\ &= \hat k \cdot \hat j - \hat k \cdot \hat k \\ &= 0 - 1 = -1 \end{aligned}
Why?

Do the cross products in the brackets first, then the dot products.

(i^×j^)⋅j^=k^⋅j^=0(j^×i^)⋅k^=(−k^)⋅k^=−(k^⋅k^)=−10+(−1)=−1\begin{aligned} (\hat i \times \hat j) \cdot \hat j &= \hat k \cdot \hat j \\ &= 0 \\[4pt] (\hat j \times \hat i) \cdot \hat k &= (-\hat k) \cdot \hat k \\ &= -(\hat k \cdot \hat k) \\ &= -1 \\[4pt] 0 + (-1) &= -1 \end{aligned}

Watch out: the minus sign from j^×i^=−k^\hat j \times \hat i = -\hat k carries through, so the last term is −1-1⁠, not +1+1⁠.

Answer: (d) −1-1

Where this comes from

Chapter 10, Vector Algebra: Products of the unit vectors i, j, k.

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