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Vector Algebra question

2002 · Shift 0 · Q92
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Vector Algebra question

2002 · Shift 0 · Q92

JEE MainMathematicsVector AlgebraMCQ+4 / −1
a→=3i^−5j^\overrightarrow a = 3\widehat i - 5\widehat ja=3i−5j​ and b→=6i^+3j^\overrightarrow b = 6\widehat i + 3\widehat jb=6i+3j​ are two vectors and c→\overrightarrow cc is a vector such that c→=a→×b→\overrightarrow c = \overrightarrow a \times \overrightarrow bc=a×b then ∣a→∣:∣b→∣:∣c→∣\left| {\overrightarrow a } \right|:\left| {\overrightarrow b } \right|:\left| {\overrightarrow c } \right|​a​:​b​:​c​ =
  1. A
    34:45:39\sqrt {34} :\sqrt {45} :\sqrt {39}34​:45​:39​
  2. B
    34:45:39\sqrt {34} :\sqrt {45} :3934​:45​:39
  3. C
    34:39:4534:39:4534:39:45
  4. D
     39:35:34\,39:35:3439:35:34
View written solutionFree

Correct answer: B

  1. Write the vectors in component form

Given, a⃗=3i^−5j^=(3,−5,0),b⃗=6i^+3j^=(6,3,0)\vec a = 3\hat i - 5\hat j = (3,-5,0), \qquad \vec b = 6\hat i + 3\hat j = (6,3,0)a=3i^−5j^​=(3,−5,0),b=6i^+3j^​=(6,3,0)

Since both are 2D vectors, we treat them as 3D vectors with zero zzz-component.

  1. Find the magnitudes of a⃗\vec aa and b⃗\vec bb

For a⃗\vec aa: ∣a⃗∣=32+(−5)2=9+25=34|\vec a| = \sqrt{3^2+(-5)^2} = \sqrt{9+25} = \sqrt{34}∣a∣=32+(−5)2​=9+25​=34​

For b⃗\vec bb: ∣b⃗∣=62+32=36+9=45|\vec b| = \sqrt{6^2+3^2} = \sqrt{36+9} = \sqrt{45}∣b∣=62+32​=36+9​=45​

  1. Compute c⃗=a⃗×b⃗\vec c = \vec a \times \vec bc=a×b

Using determinant form,

a⃗×b⃗=∣i^j^k^3−50630∣\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{vmatrix}a×b=​i^36​j^​−53​k^00​​

Expanding,

a⃗×b⃗=i^((−5)(0)−0(3))−j^(3(0)−0(6))+k^(3⋅3−(−5)⋅6)\vec a \times \vec b = \hat i\big((-5)(0)-0(3)\big) - \hat j\big(3(0)-0(6)\big) + \hat k\big(3\cdot 3-(-5)\cdot 6\big)a×b=i^((−5)(0)−0(3))−j^​(3(0)−0(6))+k^(3⋅3−(−5)⋅6) =0i^−0j^+(9+30)k^=39k^= 0\hat i - 0\hat j + (9+30)\hat k = 39\hat k=0i^−0j^​+(9+30)k^=39k^

So, c⃗=39k^\vec c = 39\hat kc=39k^

  1. Find the magnitude of c⃗\vec cc

∣c⃗∣=∣39k^∣=39|\vec c| = |39\hat k| = 39∣c∣=∣39k^∣=39

  1. Form the required ratio

∣a⃗∣:∣b⃗∣:∣c⃗∣=34:45:39|\vec a|:|\vec b|:|\vec c| = \sqrt{34}:\sqrt{45}:39∣a∣:∣b∣:∣c∣=34​:45​:39

  1. Compare with options

This matches Option B.

Derived Answer: 34:45:39\boxed{\sqrt{34}:\sqrt{45}:39}34​:45​:39​

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