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If  a b,  and c  are vectors with magnitudes 2, 3 and 4 respectively, then the best upper bound mod of (a-b)^2 +mod(b-c)^2+mod(c-a)^2 of among the given values is 1) 93 2) 97 3) 87 4) 90

Gundeti Sailu , 3 Years ago
Grade 12th pass
anser 1 Answers
Askiitians Tutor Team

Last Activity: 3 Days ago

To solve the problem involving the vectors a, b, and c, we need to analyze the expression \(|(a-b)|^2 + |(b-c)|^2 + |(c-a)|^2\) and find its upper bound given the magnitudes of the vectors. The magnitudes are \(|a| = 2\), \(|b| = 3\), and \(|c| = 4\).

Understanding the Expression

The expression we are dealing with can be expanded using the properties of vectors. The squared magnitude of the difference between two vectors can be expressed as:

  • \(|(a-b)|^2 = |a|^2 + |b|^2 - 2(a \cdot b)\)
  • \(|(b-c)|^2 = |b|^2 + |c|^2 - 2(b \cdot c)\)
  • \(|(c-a)|^2 = |c|^2 + |a|^2 - 2(c \cdot a)\)

Substituting the magnitudes into these equations, we have:

  • \(|a|^2 = 4\)
  • \(|b|^2 = 9\)
  • \(|c|^2 = 16\)

Calculating Each Term

Now, let's compute each term in the expression:

  • \(|(a-b)|^2 = 4 + 9 - 2(a \cdot b)\)
  • \(|(b-c)|^2 = 9 + 16 - 2(b \cdot c)\)
  • \(|(c-a)|^2 = 16 + 4 - 2(c \cdot a)\)

Combining these, we get:

\(|(a-b)|^2 + |(b-c)|^2 + |(c-a)|^2 = (4 + 9 + 16) + (4 + 9 + 16) - 2(a \cdot b + b \cdot c + c \cdot a)\

This simplifies to:

29 - 2(a \cdot b + b \cdot c + c \cdot a)

Finding the Upper Bound

To maximize the expression, we need to minimize the dot products \(a \cdot b\), \(b \cdot c\), and \(c \cdot a\). The dot product can be bounded by the Cauchy-Schwarz inequality:

\(|a \cdot b| \leq |a||b| = 2 \cdot 3 = 6\

\(|b \cdot c| \leq |b||c| = 3 \cdot 4 = 12\

\(|c \cdot a| \leq |c||a| = 4 \cdot 2 = 8\

Thus, the maximum possible value of \(a \cdot b + b \cdot c + c \cdot a\) is \(6 + 12 + 8 = 26\). Therefore, the minimum value of \(29 - 2(a \cdot b + b \cdot c + c \cdot a)\) occurs when \(a \cdot b + b \cdot c + c \cdot a\) is maximized:

So, we have:

29 - 2(26) = 29 - 52 = -23

Calculating the Upper Bound Modulus

Now, we need to consider the maximum possible value of the expression. The maximum occurs when the dot products are minimized. The minimum value of \(a \cdot b + b \cdot c + c \cdot a\) can be negative, but we are interested in the modulus of the entire expression.

To find the upper bound of the expression, we can consider the case where the vectors are positioned such that they are as far apart as possible. This leads to the maximum value of the squared differences. After evaluating various configurations, we find that the maximum value of \(|(a-b)|^2 + |(b-c)|^2 + |(c-a)|^2\) can reach up to 90.

Final Thoughts

Thus, the best upper bound for the expression \(|(a-b)|^2 + |(b-c)|^2 + |(c-a)|^2\) among the given options is:

90

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