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Vector Algebra question

2002 · Shift 0 · Q90
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  5. /2002 · Shift 0 · Q90

Vector Algebra question

2002 · Shift 0 · Q90

JEE MainMathematicsVector AlgebraMCQ+4 / −1
If the vectors a→,b→\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}a,b and c→\overrightarrow{\mathbf{c}}c from the sides BC,CAB C, C ABC,CA and ABA BAB respectively of a triangle ABCA B CABC, then :
  1. A
    a→⋅b→=b→⋅c→=c→⋅b→=0\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}}=0a⋅b=b⋅c=c⋅b=0
  2. B
    a→×b→=b→×c→=c→×a→\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}a×b=b×c=c×a
  3. C
    a→⋅b→=b→⋅c→=c→⋅a→=0\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}=0a⋅b=b⋅c=c⋅a=0
  4. D
    a→×a→+a→×c→+c→×a→=0→\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{0}}a×a+a×c+c×a=0
View written solutionFree

Correct answer: B, D

  1. Interpret the given vectors

Since the vectors are along the sides of triangle ABCABCABC taken in order,

a⃗=BC→,b⃗=CA→,c⃗=AB→\vec a = \overrightarrow{BC}, \qquad \vec b = \overrightarrow{CA}, \qquad \vec c = \overrightarrow{AB}a=BC,b=CA,c=AB

For any triangle, the side vectors taken cyclically satisfy

AB→+BC→+CA→=0⃗\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec 0AB+BC+CA=0

So,

c⃗+a⃗+b⃗=0⃗\vec c + \vec a + \vec b = \vec 0c+a+b=0

or

a⃗+b⃗+c⃗=0⃗\vec a+\vec b+\vec c=\vec 0a+b+c=0
  1. Check Option B

From

c⃗=−(a⃗+b⃗)\vec c = -(\vec a+\vec b)c=−(a+b)

Now,

b⃗×c⃗=b⃗×(−(a⃗+b⃗))=−b⃗×a⃗−b⃗×b⃗\vec b \times \vec c = \vec b \times (-(\vec a+\vec b)) = -\vec b\times \vec a - \vec b\times \vec bb×c=b×(−(a+b))=−b×a−b×b

Since b⃗×b⃗=0⃗\vec b\times \vec b=\vec 0b×b=0,

b⃗×c⃗=−b⃗×a⃗=a⃗×b⃗\vec b \times \vec c = -\vec b\times \vec a = \vec a\times \vec bb×c=−b×a=a×b

Similarly,

c⃗×a⃗=(−(a⃗+b⃗))×a⃗=−a⃗×a⃗−b⃗×a⃗=−b⃗×a⃗=a⃗×b⃗\vec c \times \vec a = (-(\vec a+\vec b))\times \vec a = -\vec a\times \vec a - \vec b\times \vec a = -\vec b\times \vec a = \vec a\times \vec bc×a=(−(a+b))×a=−a×a−b×a=−b×a=a×b

Hence,

a⃗×b⃗=b⃗×c⃗=c⃗×a⃗\vec a\times \vec b = \vec b\times \vec c = \vec c\times \vec aa×b=b×c=c×a

So Option B is true.

  1. Check Option A

Option A says

a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅b⃗=0\vec a\cdot \vec b=\vec b\cdot \vec c=\vec c\cdot \vec b=0a⋅b=b⋅c=c⋅b=0

This would mean the side vectors are mutually perpendicular in a way that is not true for a general triangle. So this is false.

  1. Check Option C

Option C says

a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=0\vec a\cdot \vec b=\vec b\cdot \vec c=\vec c\cdot \vec a=0a⋅b=b⋅c=c⋅a=0

Again, this would imply each pair is perpendicular, which is impossible for arbitrary side vectors of a triangle. Hence false.

  1. Check Option D

Option D is

a⃗×a⃗+a⃗×c⃗+c⃗×a⃗=0⃗\vec a\times \vec a+\vec a\times \vec c+\vec c\times \vec a=\vec 0a×a+a×c+c×a=0

Now,

a⃗×a⃗=0⃗\vec a\times \vec a=\vec 0a×a=0

and

c⃗×a⃗=−(a⃗×c⃗)\vec c\times \vec a = -(\vec a\times \vec c)c×a=−(a×c)

Therefore,

a⃗×a⃗+a⃗×c⃗+c⃗×a⃗=0⃗+a⃗×c⃗−a⃗×c⃗=0⃗\vec a\times \vec a+\vec a\times \vec c+\vec c\times \vec a = \vec 0 + \vec a\times \vec c - \vec a\times \vec c = \vec 0a×a+a×c+c×a=0+a×c−a×c=0

So Option D is also true.

  1. Conclusion

The correct statements are:

B and D\boxed{B \text{ and } D}B and D​

Since this is labeled as an MCQ (single correct), there seems to be an issue in the options/stored answer. Mathematically, both BBB and DDD are correct.

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