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Vector Algebra question

2024 · 1 Feb · Shift 2 · Q58
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Vector Algebra question

2024 · 1 Feb · Shift 2 · Q58

JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let a→=i^+j^+k^,b→=−i^−8j^+2k^\overrightarrow{\mathrm{a}}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=-\hat{i}-8 \hat{j}+2 \hat{k}a=i^+j^​+k^,b=−i^−8j^​+2k^ and c→=4i^+c2j^+c3k^\overrightarrow{\mathrm{c}}=4 \hat{i}+\mathrm{c}_2 \hat{j}+\mathrm{c}_3 \hat{k}c=4i^+c2​j^​+c3​k^ be three vectors such that b→×a→=c→×a→\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}b×a=c×a. If the angle between the vector c→\overrightarrow{\mathrm{c}}c and the vector 3i^+4j^+k^3 \hat{i}+4 \hat{j}+\hat{k}3i^+4j^​+k^ is θ\thetaθ, then the greatest integer less than or equal to tan⁡2θ\tan ^2 \thetatan2θ is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 38

  1. Given vectors

a⃗=(1,1,1),b⃗=(−1,−8,2),c⃗=(4,c2,c3)\vec a=(1,1,1),\quad \vec b=(-1,-8,2),\quad \vec c=(4,c_2,c_3)a=(1,1,1),b=(−1,−8,2),c=(4,c2​,c3​)

and

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

This implies

(b⃗−c⃗)×a⃗=0⃗.(\vec b-\vec c)\times \vec a=\vec 0.(b−c)×a=0.

So, b⃗−c⃗\vec b-\vec cb−c is parallel to a⃗\vec aa.

  1. Use parallel condition

First compute:

b⃗−c⃗=(−1−4,−8−c2,2−c3)=(−5,−8−c2,2−c3).\vec b-\vec c=(-1-4,-8-c_2,2-c_3)=(-5,-8-c_2,2-c_3).b−c=(−1−4,−8−c2​,2−c3​)=(−5,−8−c2​,2−c3​).

Since this is parallel to (1,1,1)(1,1,1)(1,1,1), we must have

(−5,−8−c2,2−c3)=λ(1,1,1).(-5,-8-c_2,2-c_3)=\lambda(1,1,1).(−5,−8−c2​,2−c3​)=λ(1,1,1).

Comparing components:

λ=−5\lambda=-5λ=−5

So,

−8−c2=−5⇒c2=−3-8-c_2=-5 \Rightarrow c_2=-3−8−c2​=−5⇒c2​=−3

and

2−c3=−5⇒c3=7.2-c_3=-5 \Rightarrow c_3=7.2−c3​=−5⇒c3​=7.

Hence,

c⃗=(4,−3,7).\vec c=(4,-3,7).c=(4,−3,7).

  1. Find angle with vector (3,4,1)\boldsymbol{(3,4,1)}(3,4,1)

Let

d⃗=(3,4,1).\vec d=(3,4,1).d=(3,4,1).

We need the angle θ\thetaθ between c⃗\vec cc and d⃗\vec dd.

Use

tan⁡2θ=∣c⃗×d⃗∣2(c⃗⋅d⃗)2.\tan^2\theta=\frac{|\vec c\times \vec d|^2}{(\vec c\cdot \vec d)^2}.tan2θ=(c⋅d)2∣c×d∣2​.

  1. Compute dot product

c⃗⋅d⃗=4⋅3+(−3)⋅4+7⋅1=12−12+7=7.\vec c\cdot \vec d=4\cdot 3+(-3)\cdot 4+7\cdot 1=12-12+7=7.c⋅d=4⋅3+(−3)⋅4+7⋅1=12−12+7=7.

So,

(c⃗⋅d⃗)2=49.(\vec c\cdot \vec d)^2=49.(c⋅d)2=49.

  1. Compute cross product
\begin{vmatrix} \hat i & \hat j & \hat k\\ 4 & -3 & 7\\ 3 & 4 & 1 \end{vmatrix}$$ $$=\hat i((-3)(1)-7\cdot 4)-\hat j(4\cdot 1-7\cdot 3)+\hat k(4\cdot 4-(-3)\cdot 3)$$ $$=\hat i(-3-28)-\hat j(4-21)+\hat k(16+9)$$ $$=(-31,17,25).$$ Thus, $$|\vec c\times \vec d|^2=(-31)^2+17^2+25^2=961+289+625=1875.$$ --- 6. **Compute** $\boldsymbol{\tan^2\theta}$ $$\tan^2\theta=\frac{1875}{49}=38+\frac{13}{49}.$$ Therefore, $$\left\lfloor \tan^2\theta\right\rfloor=38.$$ --- 7. **Comparison with stored answer** Derived answer: $38$ Stored correct answer: $38$ They match.
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