Let the Vectors u₁ = i + j + ak, u₂ = i + bj + k and u₃ = ci + j + k Be Coplanar. If v₁ = (a+b)i + cj + ck, v₂ = ai + (b+c)j + ak and v₃ = bi + bj + (c+a)k Are Also Coplanar, Then Find the Value of 6(a + b + c)

JEE Main 2023 — 10 April, Morning Shift. Previous Year Question.

Problem

Let the vectors $\vec u_1=\hat i+\hat j+a\hat k$, $\vec u_2=\hat i+b\hat j+\hat k$, and $\vec u_3=c\hat i+\hat j+\hat k$ be coplanar. If the vectors $\vec v_1=(a+b)\hat i+c\hat j+c\hat k$, $\vec v_2=a\hat i+(b+c)\hat j+a\hat k$, and $\vec v_3=b\hat i+b\hat j+(c+a)\hat k$ are also coplanar, then find the value of $6(a+b+c)$.

Key insight. “Coplanar” for three vectors always means the same thing — their scalar triple product (the $3\times3$ determinant of their components) is zero. Applying that condition twice, once to each set of vectors, produces two equations in $a$, $b$, $c$ that combine to a single clean relation.

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Approach

Set the determinant of $\vec u_1,\vec u_2,\vec u_3$’s components to zero — this is the coplanarity condition — and simplify to get a relation between $a$, $b$, $c$. Do the same for $\vec v_1,\vec v_2,\vec v_3$, which is algebraically messier but simplifies dramatically due to cancellation. Combining the two resulting equations gives $a+b+c$ directly.

Solution

Step 1 — Write the coplanarity condition for u1, u2, u3

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

Step 2 — Expand this determinant

$$1(b\cdot1-1\cdot1) – 1(1\cdot1-1\cdot c) + a(1\cdot1-b\cdot c) = 0$$

$$(b-1)-(1-c)+a(1-bc) = 0$$

$$b-1-1+c+a-abc = 0$$

$$a+b+c-abc = 2 \quad \text{…(1)}$$

Step 3 — Write the coplanarity condition for v1, v2, v3

$$\begin{vmatrix} a+b & c & c \\ a & b+c & a \\ b & b & c+a \end{vmatrix} = 0$$

Step 4 — Expand this determinant

Expanding along the first row:

$$(a+b)\big[(b+c)(c+a)-ab\big] – c\big[a(c+a)-ab\big] + c\big[ab-(b+c)b\big] = 0$$

Simplifying each bracket: $(b+c)(c+a)-ab = bc+ab+c^2+ac-ab = c(a+b+c)$; $\ a(c+a)-ab=a(a+c-b)$; $\ ab-(b+c)b=b(a-b-c)$.

$$(a+b)\cdot c(a+b+c) – c\cdot a(a+c-b) + c\cdot b(a-b-c) = 0$$

Factoring out $c$:

$$c\Big[(a+b)(a+b+c) – a(a+c-b) + b(a-b-c)\Big] = 0$$

Step 5 — Simplify the bracket

Expanding each piece: $(a+b)(a+b+c)=a^2+2ab+b^2+ac+bc$; $\ -a(a+c-b)=-a^2-ac+ab$; $\ b(a-b-c)=ab-b^2-bc$.

Adding all three: the $a^2$, $b^2$, $ac$, and $bc$ terms all cancel, leaving:

$$2ab+ab+ab = 4ab$$

So the coplanarity condition becomes:

$$c\cdot4ab = 0 \implies 4abc=0 \implies abc=0$$

Step 6 — Combine with equation (1)

From Step 5, $abc=0$. Substituting into equation (1):

$$a+b+c-0 = 2 \implies a+b+c=2$$

Step 7 — Compute 6(a+b+c)

$$6(a+b+c) = 6\times2 = 12$$

Answer

Common mistakes

• Losing track of terms while expanding the second determinant. With three variables and several product terms in each bracket, it’s easy to drop a term — grouping the expansion by which pair of variables each term involves (as in Step 5) makes the cancellation to $4ab$ easier to verify.
• Trying to solve for a, b, c individually. The problem only asks for $a+b+c$ (times 6), and the two coplanarity conditions alone don’t pin down individual values of $a$, $b$, $c$ — only their combination via equations (1) and $abc=0$. Chasing individual values is unnecessary and, with only two independent equations for three unknowns, not even fully possible.
• Forgetting that $abc=0$ substitutes cleanly into equation (1). It’s tempting to treat $abc=0$ and equation (1) as two separate unused facts rather than combining them — but equation (1) was written specifically as $a+b+c-abc=2$ so that $abc=0$ collapses it immediately.

Practise next

• Under the same two coplanarity conditions, find $(a+b+c)^2+abc$.