AP EAMCET · Maths · Vector Algebra
If the vectors \(2 \overline{\mathrm{i}}+3 \overline{\mathrm{j}}+l \overline{\mathrm{k}},-3 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}-4 l \overline{\mathrm{k}}\) and \(\overline{\mathrm{i}}-\overline{\mathrm{j}}+3 l \overline{\mathrm{k}}\) form a right angled triangle for a positive value of \(l\), then the length of its hypotenuse is
- A \(\sqrt{\frac{40}{3}}\)
- B \(\sqrt{\frac{55}{3}}\)
- C \(\sqrt{\frac{65}{3}}\)
- D \(\sqrt{\frac{59}{3}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{55}{3}}\)
Step-by-step Solution
Detailed explanation
\( \vec{A} \cdot \vec{C} = (2)(1) + (3)(-1) + (l)(3l) = 2 - 3 + 3l^2 = -1 + 3l^2 \) \( -1 + 3l^2 = 0 \Rightarrow l^2 = \frac{1}{3} \) Hypotenuse is \( \vec{B} \). Length \( |\vec{B}| = \sqrt{(-3)^2 + (-2)^2 + (-4l)^2} = \sqrt{9 + 4 + 16l^2} = \sqrt{13 + 16l^2} \)…
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