CBSE Sample Paper 2023-24 · Section A · 1 mark
Q12: Vector component of 4 i ^ + 6 j ^ 4\hat i + 6\hat j 4 i ^ + 6 j ^ 3 j ^ + 4 k ^ 3\hat j + 4\hat k 3 j ^ + 4 k ^
With chapter links Answers only
The question
If a ⃗ = 4 i ^ + 6 j ^ \vec a = 4\hat i + 6\hat j a = 4 i ^ + 6 j ^ b ⃗ = 3 j ^ + 4 k ^ \vec b = 3\hat j + 4\hat k b = 3 j ^ + 4 k ^ a ⃗ \vec a a b ⃗ \vec b b
(a) 18 5 ( 3 i ^ + 4 k ^ ) \frac{18}{5}\left(3\hat i + 4\hat k\right) 5 18 ( 3 i ^ + 4 k ^ )
(b) 18 25 ( 3 j ^ + 4 k ^ ) \frac{18}{25}\left(3\hat j + 4\hat k\right) 25 18 ( 3 j ^ + 4 k ^ )
(c) 18 5 ( 3 i ^ + 4 k ^ ) \frac{18}{5}\left(3\hat i + 4\hat k\right) 5 18 ( 3 i ^ + 4 k ^ )
(d) 18 25 ( 4 i ^ + 6 j ^ ) \frac{18}{25}\left(4\hat i + 6\hat j\right) 25 18 ( 4 i ^ + 6 j ^ )
Exam answer Show all explanations
a ⃗ ⋅ b ⃗ = 0 + 18 + 0 = 18 ∣ b ⃗ ∣ 2 = 9 + 16 = 25 \begin{aligned} \vec a \cdot \vec b &= 0 + 18 + 0 = 18 \\ |\vec b|^2 &= 9 + 16 = 25 \end{aligned} a ⋅ b ∣ b ∣ 2 = 0 + 18 + 0 = 18 = 9 + 16 = 25 Why?
Write both vectors with all three components: a ⃗ = 4 i ^ + 6 j ^ + 0 k ^ \vec a = 4\hat i + 6\hat j + 0\hat k a = 4 i ^ + 6 j ^ + 0 k ^ b ⃗ = 0 i ^ + 3 j ^ + 4 k ^ \vec b = 0\hat i + 3\hat j + 4\hat k b = 0 i ^ + 3 j ^ + 4 k ^
a ⃗ ⋅ b ⃗ = 4 ( 0 ) + 6 ( 3 ) + 0 ( 4 ) ∣ b ⃗ ∣ 2 = 0 2 + 3 2 + 4 2 \begin{aligned} \vec a \cdot \vec b &= 4(0) + 6(3) + 0(4) \\ |\vec b|^2 &= 0^2 + 3^2 + 4^2 \end{aligned} a ⋅ b ∣ b ∣ 2 = 4 ( 0 ) + 6 ( 3 ) + 0 ( 4 ) = 0 2 + 3 2 + 4 2 Vector component of a ⃗ \vec a a b ⃗ \vec b b
( a ⃗ ⋅ b ⃗ ∣ b ⃗ ∣ 2 ) b ⃗ = 18 25 ( 3 j ^ + 4 k ^ ) \begin{aligned} &\left(\frac{\vec a \cdot \vec b}{|\vec b|^2}\right)\vec b \\ &= \frac{18}{25}\left(3\hat j + 4\hat k\right) \end{aligned} ( ∣ b ∣ 2 a ⋅ b ) b = 25 18 ( 3 j ^ + 4 k ^ ) Why?
The vector component is the projection length times the unit vector along b ⃗ \vec b b
a ⃗ ⋅ b ⃗ ∣ b ⃗ ∣ ⋅ b ⃗ ∣ b ⃗ ∣ = ( a ⃗ ⋅ b ⃗ ∣ b ⃗ ∣ 2 ) b ⃗ \begin{aligned} &\frac{\vec a \cdot \vec b}{|\vec b|} \cdot \frac{\vec b}{|\vec b|} \\ &= \left(\frac{\vec a \cdot \vec b}{|\vec b|^2}\right)\vec b \end{aligned} ∣ b ∣ a ⋅ b ⋅ ∣ b ∣ b = ( ∣ b ∣ 2 a ⋅ b ) b Here ∣ b ⃗ ∣ = 5 |\vec b| = 5 ∣ b ∣ = 5 18 5 \frac{18}{5} 5 18 1 5 ( 3 j ^ + 4 k ^ ) \frac{1}{5}\left(3\hat j + 4\hat k\right) 5 1 ( 3 j ^ + 4 k ^ )
∣ b ⃗ ∣ = 25 = 5 18 5 ⋅ 1 5 ( 3 j ^ + 4 k ^ ) = 18 5 × 5 ( 3 j ^ + 4 k ^ ) = 18 25 ( 3 j ^ + 4 k ^ ) \begin{aligned} |\vec b| &= \sqrt{25} = 5 \\[4pt] &\frac{18}{5} \cdot \frac{1}{5}\left(3\hat j + 4\hat k\right) \\ &= \frac{18}{5 \times 5}\left(3\hat j + 4\hat k\right) \\ &= \frac{18}{25}\left(3\hat j + 4\hat k\right) \end{aligned} ∣ b ∣ = 25 = 5 5 18 ⋅ 5 1 ( 3 j ^ + 4 k ^ ) = 5 × 5 18 ( 3 j ^ + 4 k ^ ) = 25 18 ( 3 j ^ + 4 k ^ ) Options (a), (c) and (d) point along other vectors, so they are ruled out.
Why?
Watch out: 18 5 \frac{18}{5} 5 18 b ⃗ = 3 j ^ + 4 k ^ \vec b = 3\hat j + 4\hat k b = 3 j ^ + 4 k ^ 3 i ^ + 4 k ^ 3\hat i + 4\hat k 3 i ^ + 4 k ^ a ⃗ \vec a a
Answer: (b) 18 25 ( 3 j ^ + 4 k ^ ) \dfrac{18}{25}\left(3\hat j + 4\hat k\right) 25 18 ( 3 j ^ + 4 k ^ )
As printed, options (a) and (c) are identical.
Where this comes from Chapter 10, Vector Algebra : Vector component (projection) of a along b.
Question type: Projection
How to spot it: click to see Type 6. Projection: the shadow of one vector on another
Spot it: "find the projection of the vector a ⃗ \vec a a b ⃗ \vec b b
Here: Here the vector component is asked, not just the length. a ⃗ ⋅ b ⃗ = 18 \vec a\cdot\vec b = 18 a ⋅ b = 18 ∣ b ⃗ ∣ = 5 |\vec b| = 5 ∣ b ∣ = 5 18 5 \frac{18}{5} 5 18 3 j ^ + 4 k ^ 5 \frac{3\hat j + 4\hat k}{5} 5 3 j ^ + 4 k ^
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