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Vector Algebra question

2021 · 25 Jul · Shift 2 · Q31

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JEE Main
Mathematics
Vector Algebra
2021 · 25 Jul · Shift 2 · Q31

Vector Algebra question

2021 · 25 Jul · Shift 2 · Q31

JEE MainMathematicsVector AlgebraMCQ+4 / −1

Let a, b and c be distinct positive numbers. If the vectors

a\widehat i + a\widehat j + c\widehat k,\widehat i+\widehat k

and

c\widehat i + c\widehat j + b\widehat k

are co-planar, then c is equal to :

A
${2 \over {{1 \over a} + {1 \over b}}}$
B
${{a + b} \over 2}$
C
${1 \over a} + {1 \over b}$
D
$\sqrt {ab}$

View written solutionFree

Correct answer: D

Write the three vectors explicitly

Let

\vec v_1=(a,a,c),\qquad \vec v_2=(1,0,1),\qquad \vec v_3=(c,c,b).

For three vectors to be coplanar, their scalar triple product must be zero:

\begin{vmatrix} a & a & c\\ 1 & 0 & 1\\ c & c & b \end{vmatrix}=0.

Compute the determinant

Expanding along the first row:

\begin{aligned} \det &= a\begin{vmatrix}0 & 1\\ c & b\end{vmatrix} - a\begin{vmatrix}1 & 1\\ c & b\end{vmatrix} + c\begin{vmatrix}1 & 0\\ c & c\end{vmatrix} \\ &= a(0\cdot b-1\cdot c)-a(1\cdot b-1\cdot c)+c(1\cdot c-0\cdot c) \\ &= a(-c)-a(b-c)+c^2 \\ &= -ac-ab+ac+c^2 \\ &= c^2-ab. \end{aligned}

Since the vectors are coplanar,

c^2-ab=0.

So,

c^2=ab.

Use positivity of numbers

Given $a,b,c$ are positive,

c=\sqrt{ab}.

Check the options

A: $\dfrac{2}{\frac1a+\frac1b}$ = harmonic mean, not necessarily $\sqrt{ab}$
B: $\dfrac{a+b}{2}$ = arithmetic mean, not necessarily $\sqrt{ab}$
C: $\dfrac1a+\dfrac1b$ is unrelated
D: $\sqrt{ab}$ ✅

Hence the correct option is D.

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