CBSE Board Examination 2023 · Section A · 1 mark

Q10: $|\vec b|$ when $\vec a + \vec b = \hat i$ and $\vec a = 2\hat i - 2\hat j + 2\hat k$

The question

If $\vec a + \vec b = \hat i$ and $\vec a = 2\hat i - 2\hat j + 2\hat k$ ⁠, then $|\vec b|$ equals :

(a) $\sqrt{14}$

(b) $3$

(d) $\sqrt{17}$

Exam answer

\begin{aligned} \vec b &= \hat i - \vec a \\ &= \hat i - (2\hat i - 2\hat j + 2\hat k) \\ &= -\hat i + 2\hat j - 2\hat k \end{aligned}

Why?

From $\vec a + \vec b = \hat i$ ⁠, subtract $\vec a$ from both sides.

Then subtract component by component: $1 - 2 = -1$ for $\hat i$ ⁠, $0 - (-2) = 2$ for $\hat j$ and $0 - 2 = -2$ for $\hat k$ ⁠.

\begin{aligned} &\hat i - (2\hat i - 2\hat j + 2\hat k) \\ &= \hat i - 2\hat i + 2\hat j - 2\hat k \\ &= (1 - 2)\hat i + 2\hat j - 2\hat k \\ &= -\hat i + 2\hat j - 2\hat k \end{aligned}

Watch out: the bracket changes every sign inside it.

\begin{aligned} |\vec b| &= \sqrt{(-1)^2 + 2^2 + (-2)^2} \\ &= \sqrt{1 + 4 + 4} = 3 \end{aligned}

Why?

The magnitude of $x\hat i + y\hat j + z\hat k$ is

\sqrt{x^2 + y^2 + z^2}

Squaring removes the minus signs, and $\sqrt{9} = 3$ ⁠.

\begin{aligned} (-1)^2 &= (-1)(-1) = 1 \\ 2^2 &= 4 \\ (-2)^2 &= (-2)(-2) = 4 \\[4pt] 1 + 4 + 4 &= 9 \\ \sqrt{9} &= 3 \end{aligned}

Answer: (b) $3$

Where this comes from

Chapter 10, Vector Algebra: Magnitude of a vector.

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Vector Algebra in past papers: question 16 of 48 · see them all

Q10: ∣b⃗∣|\vec b|∣b∣ when a⃗+b⃗=i^\vec a + \vec b = \hat ia+b=i^ and a⃗=2i^−2j^+2k^\vec a = 2\hat i - 2\hat j + 2\hat ka=2i^−2j^​+2k^