Q.If $\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}, \vec{b}=-\hat{i}+2\hat{j}+\hat{k}$ and $\vec{c}=3\hat{i}+\hat{j}$ are such that $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c},$ then find the value of $\lambda.$

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.The set of real values of λ for which the vectors $\lambda\vec{i}-3\vec{j}+5\vec{k}$ and $2\lambda\vec{i}-\lambda\vec{j}+\vec{k}$ are perpendicular to each other is (A) $\{0, 1\}$ (B) $\{-2\}$ (C) $\{2, -1\}$ (D) $\varphi$

   ›Reveal solutionSolution

   Perpendicular vectors have zero dot product; the resulting quadratic in $\lambda$ has no real roots, so the answer set is empty.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product vanishes. Setting up that equation converts a geometry condition into an algebraic one in $\lambda$.

   Step-by-Step Solution

   1. The vectors are $\vec u=(\lambda,-3,5)$ and $\vec v=(2\lambda,-\lambda,1)$.
   2. Perpendicularity: $\vec u\cdot\vec v=0$: $\ \lambda(2\lambda)+(-3)(-\lambda)+5(1)=0$.
   3. Simplify: $2\lambda^2+3\lambda+5=0$.
   4. Discriminant $=3^2-4(2)(5)=9-40=-31<0$. …

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2. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.$\bar{a} = \bar{i} - \bar{j} + \bar{k}$, $\bar{b} = 2\bar{i} + \bar{j} + \bar{k}$ are two vectors and $\bar{c}$ is a unit vector lying in the plane of $\bar{a}$ and $\bar{b}$. If $\bar{c}$ is perpendicular to $\bar{b}$ then $\bar{c}.(\bar{i}+\bar{j}+2\bar{k}) =$ (A) $0$ (B) $5$ (C) $\dfrac{1}{\sqrt{21}}$ (D) $\dfrac{2}{\sqrt{21}}$

   ›Reveal solutionSolution

   This tests finding a unit vector coplanar with two given vectors and perpendicular to one of them; the required dot product works out to $\dfrac{1}{\sqrt{21}}$.

   Concept and Intuition

   Any vector in the plane spanned by $\bar a$ and $\bar b$ can be written as a linear combination $m\bar a+n\bar b$. Imposing perpendicularity to $\bar b$ gives one linear equation in $m,n$, pinning down the direction of $\bar c$ up to a scalar (which is then fixed by the unit-length condition).

   Step-by-Step Solution

   1. Let $\bar c=m\bar a+n\bar b$ where $\bar a=(1,-1,1),\ \bar b=(2,1,1)$.
   2. $\bar c\cdot\bar b=0\Rightarrow m(\bar a\cdot\bar b)+n(\bar b\cdot\bar b)=0$.
   3. $\bar a\cdot\bar b=1(2)+(-1)(1)+1(1)=2-1+1=2$. $\bar b\cdot\bar b=4+1+1=6$.
   4. So $2m+6n=0\Rightarrow m=-3n$.
   5. $\bar c\parallel -3n\bar a+n\bar b=n(-3\bar a+\bar b)=n\big((-3,3,-3)+(2,1,1)\big)=n(-1,4,-2)$.
   6. Direction vector $(-1,4,-2)$ has magnitude $\sqrt{1+16+4}=\sqrt{21}$, so the unit vector is $\pm\dfrac{(-1,4,-2)}{\sqrt{21}}$. …

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3. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $\bar{a}=2\bar{i}+3\bar{j}, \bar{b}=3\bar{j}+4\bar{k}$ and $\bar{c}=5\bar{i}+4\bar{k}$ are three vectors, then a vector which is perpendicular to $\bar{a}$ and $\bar{b}\times\bar{c}$ is (A) $45\bar{i}-30\bar{j}+15\bar{k}$ (B) $3\bar{i}-2\bar{j}+\bar{k}$ (C) $-30\bar{i}+20\bar{j}+4\bar{k}$ (D) $-45\bar{i}+30\bar{j}+4\bar{k}$

   ›Reveal solutionSolution

   This tests the vector-triple-product idea: a vector perpendicular to both $\bar a$ and $\bar b\times\bar c$ is simply $\bar a\times(\bar b\times\bar c)$.

   Concept and Intuition

   The cross product of any two vectors is perpendicular to both of them. So if we want a single vector perpendicular to $\bar a$ AND to $\bar b\times\bar c$, the natural candidate is $\bar a\times(\bar b\times\bar c)$ — it is perpendicular to $\bar a$ by definition of cross product, and perpendicular to $\bar b\times\bar c$ for the same reason. No need to invoke the full triple-product expansion formula; we just compute it directly.

   Step-by-Step Solution

   1. Given $\bar a=2\bar i+3\bar j+0\bar k$, $\bar b=0\bar i+3\bar j+4\bar k$, $\bar c=5\bar i+0\bar j+4\bar k$.
   2. Compute $\bar b\times\bar c=\begin{vmatrix}\bar i&\bar j&\bar k\\0&3&4\\5&0&4\end{vmatrix}$ $=\bar i(3\cdot4-4\cdot0)-\bar j(0\cdot4-4\cdot5)+\bar k(0\cdot0-3\cdot5)=12\bar i+20\bar j-15\bar k$.
   3. Compute $\bar a\times(\bar b\times\bar c)=\begin{vmatrix}\bar i&\bar j&\bar k\\2&3&0\\12&20&-15\end{vmatrix}$ …

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4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.$\bar{a},\bar{b},\bar{c}$ are unit vectors. If $\bar{a},\bar{b}$ are perpendicular vectors, $(\bar{a}-\bar{c}).(\bar{b}+\bar{c})=0$ and $\bar{c}=l\bar{a}+m\bar{b}+n(\bar{a}\times\bar{b})$; (l, m, n are scalars), then $n^2=$ (A) $l^2+m^2$ (B) $-2lm$ (C) $2l-2m$ (D) $lm+l+m$

   ›Reveal solutionSolution

   Because $\bar a,\bar b,\bar a\times\bar b$ form an orthonormal triad, decomposing $\bar c$ in this basis and using the given perpendicularity condition shows $n^2=-2lm$.

   Concept and Intuition

   When $\bar a$ and $\bar b$ are perpendicular unit vectors, $\bar a\times\bar b$ is automatically a unit vector too (since $|\bar a\times\bar b|=|\bar a||\bar b|\sin 90°=1$) and is perpendicular to both $\bar a$ and $\bar b$. So $\{\bar a,\bar b,\bar a\times\bar b\}$ is an orthonormal basis — any vector's components along these three directions are just its dot products with each, and its squared magnitude is simply the sum of squared components (Pythagoras in 3D).

   Step-by-Step Solution

   1. Since $\bar a\perp\bar b$ and both are unit vectors, $\bar a.\bar b=0$ and $\{\bar a,\bar b,\bar a\times\bar b\}$ is orthonormal.
   2. Expand $(\bar a-\bar c).(\bar b+\bar c)=0$: $\bar a.\bar b+\bar a.\bar c-\bar c.\bar b-\bar c.\bar c=0$.
   3. Since $\bar a.\bar b=0$ and $\bar c.\bar c=|\bar c|^2=1$ (unit vector): $\bar a.\bar c-\bar b.\bar c-1=0 \Rightarrow \bar a.\bar c-\bar b.\bar c=1$.
   4. Given $\bar c=l\bar a+m\bar b+n(\bar a\times\bar b)$, dot with $\bar a$: $\bar a.\bar c=l(\bar a.\bar a)+m(\bar a.\bar b)+n\cdot\bar a.(\bar a\times\bar b)=l(1)+m(0)+n(0)=l$ (since $\bar a.(\bar a\times\bar b)=0$, a vector is always perpendicular to a cross product it's part of).
   5. Similarly, dot with $\bar b$: $\bar b.\bar c=l(\bar b.\bar a)+m(\bar b.\bar b)+n\cdot\bar b.(\bar a\times\bar b)=0+m(1)+0=m$.
   6. From step 3: $l-m=1$. …

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5. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $\vec a$ and $\vec b$ are two vectors such that $|\vec a| = 2$, $|\vec b| = 3$ and $\vec a + t\vec b$ and $\vec a - t\vec b$ are perpendicular, where 't' is a positive scalar, then (A) $t = \pm\dfrac{2}{3}$ (B) $t = \dfrac{4}{9}$ (C) $t = \dfrac{2}{3}$ (D) $t = \dfrac{2}{9}$

   ›Reveal solutionSolution

   Perpendicularity of $\vec a+t\vec b$ and $\vec a-t\vec b$ forces $|\vec a|^2=t^2|\vec b|^2$, giving the positive value $t=2/3$.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product is zero. Expanding $(\vec a+t\vec b)\cdot(\vec a-t\vec b)$ using the distributive property of the dot product collapses to a simple difference of squared magnitudes, since $\vec a\cdot\vec b$ cancels.

   Step-by-Step Solution

   1. $(\vec a+t\vec b)\cdot(\vec a-t\vec b) = \vec a\cdot\vec a - t\,\vec a\cdot\vec b + t\,\vec b\cdot\vec a - t^2\,\vec b\cdot\vec b = |\vec a|^2 - t^2|\vec b|^2$.
   2. Setting this to zero (perpendicularity): $|\vec a|^2 = t^2|\vec b|^2$.
   3. Substitute $|\vec a|=2$, $|\vec b|=3$: $4 = 9t^2 \Rightarrow t^2=\dfrac49$. …

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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The vector of magnitude 2 lying in the plane of $\bar{a} = 2\bar{i} - \bar{j} + \bar{k}$ and $\bar{b} = \bar{i} + 3\bar{j} - 5\bar{k}$ and perpendicular to the vector $\bar{c} = \bar{i} + \bar{j} + \bar{k}$ is (A) $\dfrac{\sqrt{2}}{\sqrt{61}}(4\bar{i}+5\bar{j}-9\bar{k})$ (B) $\dfrac{\sqrt{2}}{\sqrt{9}}(2\bar{i}+3\bar{j}-5\bar{k})$ (C) $\dfrac{\sqrt{2}}{\sqrt{31}}(\bar{i}+5\bar{j}-6\bar{k})$ (D) $\dfrac{\sqrt{2}}{\sqrt{13}}(-\bar{i}-3\bar{j}+4\bar{k})$

   ›Reveal solutionSolution

   This tests writing a vector "in the plane of $\bar a,\bar b$" as a linear combination $\alpha\bar a+\beta\bar b$, using perpendicularity to $\bar c$ to pin the ratio $\alpha:\beta$, and finally scaling the resulting direction to the required magnitude.

   Concept and Intuition

   Every vector lying in the plane spanned by $\bar a$ and $\bar b$ is some linear combination $\alpha\bar a+\beta\bar b$ — that's what "lying in the plane" means. The extra condition (perpendicular to $\bar c$) gives one linear equation in $\alpha,\beta$, which fixes their ratio (the direction is determined up to an overall scale). The magnitude condition then fixes that scale.

   Step-by-Step Solution

   1. Let $\bar d=\alpha\bar a+\beta\bar b$ for some scalars $\alpha,\beta$ (this covers every vector in the plane of $\bar a,\bar b$).
   2. Require $\bar d\cdot\bar c=0$: $\alpha(\bar a\cdot\bar c)+\beta(\bar b\cdot\bar c)=0$.
   3. $\bar a\cdot\bar c=(2)(1)+(-1)(1)+(1)(1)=2-1+1=2$. $\bar b\cdot\bar c=(1)(1)+(3)(1)+(-5)(1)=1+3-5=-1$.
   4. So $2\alpha-\beta=0\Rightarrow\beta=2\alpha$. Taking $\alpha=1,\beta=2$: direction $=\bar a+2\bar b=(2+2,\,-1+6,\,1-10)=(4,5,-9)$.
   5. Magnitude of this direction: $\sqrt{4^2+5^2+(-9)^2}=\sqrt{16+25+81}=\sqrt{122}$. …

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7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A unit vector that is perpendicular to the vector $2\bar{i} - \bar{j} + 2\bar{k}$ and coplanar with the vectors $\bar{i} + \bar{j} - \bar{k}$ and $2\bar{i} + 2\bar{j} - \bar{k}$ is (A) $\dfrac{\bar{i} + 2\bar{j} + \bar{k}}{\sqrt{6}}$ (B) $\dfrac{3\bar{i} + 2\bar{j} - 2\bar{k}}{\sqrt{17}}$ (C) $\dfrac{2\bar{i} + 2\bar{j} - \bar{k}}{3}$ (D) $\dfrac{3\bar{i} + 2\bar{j} + 2\bar{k}}{\sqrt{17}}$

   ›Reveal solutionSolution

   Write the general coplanar combination of the two given vectors as $a\,p+b\,q$, impose perpendicularity to the third vector to pin down $a=0$, then normalize the resulting direction.

   Concept and Intuition

   "Coplanar with $p$ and $q$" means the target vector is some linear combination $a\,p+b\,q$ (this spans exactly the plane through the origin containing both). Imposing perpendicularity to a third given vector is then just one linear equation in $a,b$ — it typically forces a ratio (or here, forces one coefficient to vanish entirely), collapsing the family to a single direction, which we then normalize to a unit vector.

   Step-by-Step Solution

   1. General coplanar vector: $v=a(\bar i+\bar j-\bar k)+b(2\bar i+2\bar j-\bar k)=(a+2b)\bar i+(a+2b)\bar j+(-a-b)\bar k$.
   2. Require $v\perp(2\bar i-\bar j+2\bar k)$: $2(a+2b)-1(a+2b)+2(-a-b)=0$.
   3. Simplify: $(a+2b)(2-1)+2(-a-b)=(a+2b)-2a-2b=-a$.
   4. So the condition reduces to $-a=0\Rightarrow a=0$.
   5. With $a=0$: $v=b(2\bar i+2\bar j-\bar k)$, i.e. $v$ is parallel to $2\bar i+2\bar j-\bar k$.
   6. Magnitude of $2\bar i+2\bar j-\bar k$ is $\sqrt{4+4+1}=3$. …

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8. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Let $\bar{a} = 3\bar{i} - \bar{j} - \bar{k}$, $\bar{b} = \bar{i} + \bar{j} - 2\bar{k}$ and $\bar{c} = 2\bar{i} + 2\bar{j} + \bar{k}$. Let $\bar{d}$ be a vector such that $|\bar{d}| = \sqrt{2}$ units. If the vector $\bar{d}$ is coplanar with $\bar{a}, \bar{b}$ and perpendicular to $\bar{c}$, then $\bar{d} =$ (A) $\pm\dfrac{1}{5}(3\bar{i} - 5\bar{j} + 4\bar{k})$ (B) $\pm\dfrac{1}{5}(-4\bar{i} + 5\bar{j} - 3\bar{k})$ (C) $\pm\dfrac{1}{5}(3\bar{i} + 5\bar{j} - 4\bar{k})$ (D) $\pm\dfrac{1}{5}(-3\bar{i} + 5\bar{j} + 4\bar{k})$

   ›Reveal solutionSolution

   $\bar d$ coplanar with $\bar a,\bar b$ means $\bar d=x\bar a+y\bar b$; perpendicularity to $\bar c$ fixes the ratio $x:y$; the given magnitude fixes the scale. The answer is (A).

   Concept and Intuition

   "Coplanar with $\bar a,\bar b$" means $\bar d$ lies in the plane spanned by $\bar a$ and $\bar b$, so it can be written as a linear combination $\bar d=x\bar a+y\bar b$ for some scalars $x,y$ (this is exactly what "coplanar with two given vectors, through the origin" means). The perpendicularity condition $\bar d\cdot\bar c=0$ then gives one constraint relating $x$ and $y$, so $\bar d$ is pinned down up to a single scalar multiple — which the given magnitude $|\bar d|=\sqrt2$ finally fixes (up to sign, since both directions along that line satisfy all the stated conditions).

   Step-by-Step Solution

   1. Given $\bar a=(3,-1,-1)$, $\bar b=(1,1,-2)$, $\bar c=(2,2,1)$.
   2. Since $\bar d$ is coplanar with $\bar a,\bar b$, write $\bar d = x\bar a + y\bar b = (3x+y,\,-x+y,\,-x-2y)$.
   3. Perpendicularity to $\bar c$: $\bar d\cdot\bar c=0$:

   $2(3x+y)+2(-x+y)+1(-x-2y)=0$

   $6x+2y-2x+2y-x-2y=0 \implies 3x+2y=0 \implies y=-\tfrac{3x}{2}.$

   4. Substitute back:

   $\bar d = \left(3x-\tfrac{3x}{2},\ -x-\tfrac{3x}{2},\ -x+3x\right) = \left(\tfrac{3x}{2},\,-\tfrac{5x}{2},\,2x\right).$

   Let $x=2t$ to clear fractions: $\bar d = (3t,-5t,4t) = t(3,-5,4)$. …

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9. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If the equation of the plane passing through the points $(1,-3,2)$, $(-2,3,1)$ and perpendicular to the plane $x + 2y - 3z = 0$ is $ax + by + cz + d = 0$, then $\dfrac{a+b}{c+d} =$ (A) $\dfrac{3}{11}$ (B) $13$ (C) $\dfrac{13}{11}$ (D) $3$

   ›Reveal solutionSolution

   This tests finding a plane through two points and perpendicular to another plane, using the cross product of the connecting direction vector and the given plane's normal. Answer: $13$.

   Concept and Intuition

   A plane's normal vector must be perpendicular to every direction lying in the plane. Since the plane contains points $P_1(1,-3,2)$ and $P_2(-2,3,1)$, the vector $\vec{P_1P_2}$ lies in the plane, so the required normal $\vec n=(a,b,c)$ satisfies $\vec n\cdot\vec{P_1P_2}=0$. Also, "perpendicular to the plane $x+2y-3z=0$" means the two planes' normals are perpendicular, so $\vec n\cdot(1,2,-3)=0$. A vector perpendicular to both $\vec{P_1P_2}$ and $(1,2,-3)$ is simply their cross product.

   Step-by-Step Solution

   1. Direction vector: $\vec d = P_2-P_1 = (-2-1,\,3-(-3),\,1-2)=(-3,6,-1)$.
   2. Normal of given plane: $\vec n_2=(1,2,-3)$.
   3. Required normal: $\vec n=\vec d\times \vec n_2 = \begin{vmatrix}\vec i&\vec j&\vec k\\-3&6&-1\\1&2&-3\end{vmatrix}$ $= \vec i(6(-3)-(-1)(2)) - \vec j((-3)(-3)-(-1)(1)) + \vec k((-3)(2)-6(1))$ $= \vec i(-18+2) - \vec j(9+1) + \vec k(-6-6) = (-16,-10,-12)$. …

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10. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.If the vectors $2\bar{i} + 3\bar{j} + l\bar{k}$, $-3\bar{i} - 2\bar{j} - 4l\bar{k}$ and $\bar{i} - \bar{j} + 3l\bar{k}$ form a right angled triangle for a positive value of $l$, then the length of its hypotenuse is (A) $\sqrt{\dfrac{40}{3}}$ (B) $\sqrt{\dfrac{55}{3}}$ (C) $\sqrt{\dfrac{65}{3}}$ (D) $\sqrt{\dfrac{59}{3}}$

    ›Reveal solutionSolution

    Because the three given vectors sum to zero, they are the side vectors of a closed triangle; finding which pair is mutually perpendicular locates the right angle, and the third side (opposite that angle) is the hypotenuse whose length we compute.

    Concept and Intuition

    If three vectors $u,v,w$ satisfy $u+v+w=\bar0$, they can be laid tip-to-tail to close a triangle — this is exactly the vector-polygon condition. The vertex where two of them (as drawn, not reversed) are mutually perpendicular is the right-angle vertex of the triangle, and the side "opposite" that vertex — i.e. the third vector — is the hypotenuse. So the whole problem reduces to (a) finding which pair dots to zero for some positive $l$, and (b) computing that third vector's magnitude.

    Step-by-Step Solution

    1. Let $u=(2,3,l)$, $v=(-3,-2,-4l)$, $w=(1,-1,3l)$.
    2. Check closure: $u+v+w=(2-3+1,\,3-2-1,\,l-4l+3l)=(0,0,0)$ — confirmed, they form a triangle.
    3. Test each pair's dot product for a value making it zero (this locates the right angle):
       • $u\cdot v = -6-6-4l^2=-12-4l^2$ — never zero for real $l$.
       • $v\cdot w = -3+2-12l^2=-1-12l^2$ — never zero for real $l$.
       • $u\cdot w = 2-3+3l^2=3l^2-1$ — zero when $l^2=\dfrac13$, i.e. $l=\dfrac{1}{\sqrt3}>0$. ✓ (matches "positive value of $l$" in the problem.) …

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11. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.Let $u = 2i + 3j + k$, $v = -3i + 2j$ and $w = i - j + 4k$. Then which of the following statement is true? (A) $u$ is perpendicular to $v$ but not $w$ (B) $v$ is perpendicular to $w$ but not $u$ (C) $w$ is perpendicular to $u$ but not $v$ (D) $u$ is perpendicular to both $v$ and $w$

    ›Reveal solutionSolution

    Direct dot products show $u\cdot v=0$ (perpendicular) and $u\cdot w=3\ne0$ (not perpendicular), matching option (A).

    Concept and Intuition

    Two vectors are perpendicular exactly when their dot product is zero. Checking each pair's dot product directly settles every option — no need for angle or cross-product computation.

    Step-by-Step Solution

    1. $u\cdot v = (2)(-3)+(3)(2)+(1)(0) = -6+6+0 = 0$ — so $u\perp v$.
    2. $u\cdot w = (2)(1)+(3)(-1)+(1)(4) = 2-3+4 = 3 \ne 0$ — so $u$ is not perpendicular to $w$.
    3. For completeness, $v\cdot w=(-3)(1)+(2)(-1)+(0)(4)=-3-2+0=-5\ne0$ — $v$ is also not perpendicular to $w$. …

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12. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.Let ABC be an equilateral triangle of side $a$. M and N are two points on the sides AB and AC respectively such that $\overline{AN} = K\overline{AC}$ and $\overline{AB} = 3\overline{AM}$. If the vectors $\overline{BN}$ and $\overline{CM}$ are perpendicular, then $K =$ (A) $\dfrac{1}{5}$ (B) $\dfrac{2}{5}$ (C) $-\dfrac{1}{5}$ (D) $-\dfrac{2}{5}$

    ›Reveal solutionSolution

    Express $\overline{BN}$ and $\overline{CM}$ in terms of the two sides from $A$, use the $60^\circ$ dot product of an equilateral triangle, and set the perpendicularity condition to zero to solve for $K=\dfrac15$.

    Concept and Intuition

    Placing the vertex $A$ at the origin turns every other point into a simple scalar multiple of the two side vectors $\overline{AB}$ and $\overline{AC}$. Perpendicularity of two vectors becomes an algebraic condition: their dot product is zero. For an equilateral triangle, $\overline{AB}.\overline{AC}=a^2\cos 60^\circ=\dfrac{a^2}{2}$.

    Step-by-Step Solution

    1. Let $A$ be the origin, $\bar c=\overline{AB}$, $\bar b=\overline{AC}$, with $|\bar b|=|\bar c|=a$ and $\bar b.\bar c=\dfrac{a^2}{2}$.
    2. Since $\overline{AB}=3\overline{AM}$, $M=\dfrac{\bar c}{3}$. Since $\overline{AN}=K\overline{AC}$, $N=K\bar b$.
    3. $\overline{BN}=N-B=K\bar b-\bar c$, and $\overline{CM}=M-C=\dfrac{\bar c}{3}-\bar b$.
    4. Perpendicularity: $\overline{BN}.\overline{CM}=0$: $(K\bar b-\bar c).\left(\dfrac{\bar c}{3}-\bar b\right)= \dfrac{K}{3}(\bar b.\bar c) - K|\bar b|^2 - \dfrac13|\bar c|^2 + \bar b.\bar c = 0$.
    5. Substitute $|\bar b|^2=|\bar c|^2=a^2$, $\bar b.\bar c=a^2/2$: …

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