Q.The value of $\hat{i}\cdot(\hat{j}\times\hat{k})+\hat{j}\cdot(\hat{i}\times\hat{k})+\hat{k}\cdot(\hat{i}\times\hat{j})$ is (A) 0 (B) -1 (C) 1 (D) 3

Showing the 12 most recent of 40 on this concept.

1. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $\vec r=x\hat i+y\hat j+z\hat k$ then $(\vec r\times\hat i)\cdot(\vec r\times\hat j)+xy=$ (A) 0 (B) 1 (C) 2 (D) 3

   ›Reveal solutionSolution

   Direct computation of both cross products and their dot product shows $(\vec r\times\hat i)\cdot(\vec r\times\hat j)=-xy$, so adding $xy$ gives exactly $0$.

   Concept and Intuition

   When a vector identity involves cross products with the coordinate unit vectors, it's usually fastest to just expand everything in components directly rather than reach for a general vector identity — the coordinate structure makes the algebra short.

   Step-by-Step Solution

   1. $\vec r = (x,y,z)$.
   2. $\vec r\times\hat i = (x,y,z)\times(1,0,0) = (y\cdot0-z\cdot0,\ z\cdot1-x\cdot0,\ x\cdot0-y\cdot1) = (0, z, -y)$.
   3. $\vec r\times\hat j = (x,y,z)\times(0,1,0) = (y\cdot0-z\cdot1,\ z\cdot0-x\cdot0,\ x\cdot1-y\cdot0) = (-z, 0, x)$. …

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2. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\bar a=2\bar i-\bar j+3\bar k$, $\bar b=-3\bar i+5\bar j-4\bar k$ and $\bar c=6\bar i-4\bar j+5\bar k$, then $(\bar a\times\bar b).(\bar b\times\bar c) =$ (A) $-216$ (B) $243$ (C) $81$ (D) $-27$

   ›Reveal solutionSolution

   Apply the Lagrange-type identity $(\bar a\times\bar b)\cdot(\bar b\times\bar c)=(\bar a\cdot\bar b)(\bar b\cdot\bar c)-(\bar a\cdot\bar c)(\bar b\cdot\bar b)$ to avoid computing two cross products directly.

   Concept and Intuition

   For vectors, $(\bar p\times\bar q)\cdot(\bar r\times\bar s)=(\bar p\cdot\bar r)(\bar q\cdot\bar s)-(\bar p\cdot\bar s)(\bar q\cdot\bar r)$. Setting $\bar p=\bar a,\bar q=\bar b,\bar r=\bar b,\bar s=\bar c$ turns a cross-product computation into simple dot products.

   Step-by-Step Solution

   1. $\bar a\cdot\bar b=(2)(-3)+(-1)(5)+(3)(-4)=-6-5-12=-23$.
   2. $\bar b\cdot\bar c=(-3)(6)+(5)(-4)+(-4)(5)=-18-20-20=-58$.
   3. $\bar a\cdot\bar c=(2)(6)+(-1)(-4)+(3)(5)=12+4+15=31$.
   4. $\bar b\cdot\bar b=9+25+16=50$. …

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3. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $\vec a,\vec b,\vec c$ are 3 vectors then $[\vec a+\vec b\ \ \vec b+\vec c\ \ \vec c+\vec a]=$ (A) $[\vec a\ \vec b\ \vec c]$ (B) 0 (C) $2[\vec a\ \vec b\ \vec c]$ (D) $3[\vec a\ \vec b\ \vec c]$

   ›Reveal solutionSolution

   This is the standard identity $[\vec a+\vec b,\vec b+\vec c,\vec c+\vec a] = 2[\vec a\ \vec b\ \vec c]$.

   Concept and Intuition

   The scalar triple product $[\vec x\ \vec y\ \vec z]=\vec x\cdot(\vec y\times\vec z)$ is linear in each of its three slots, and it vanishes whenever any two of its arguments are equal (since the vectors would then be coplanar/repeated). Expanding a sum-of-vectors triple product using this multilinearity, most of the resulting eight terms vanish, and what remains combines to a clean multiple of $[\vec a\ \vec b\ \vec c]$.

   Step-by-Step Solution

   1. Expand $[\vec a+\vec b,\ \vec b+\vec c,\ \vec c+\vec a]$ using linearity in each slot — this produces 8 triple products of the form $[\vec p\ \vec q\ \vec r]$ where each of $p,q,r\in\{\vec a,\vec b,\vec c\}$ (with repeats allowed).
   2. Any triple product with a repeated vector (e.g. $[\vec a\ \vec a\ \vec c]$, $[\vec b\ \vec b\ \vec a]$) is zero.
   3. The surviving distinct terms are $[\vec a\ \vec b\ \vec c]$ and $[\vec b\ \vec c\ \vec a]$ (equal, since a cyclic permutation of a triple product leaves it unchanged), plus similar cyclic repeats — collecting them all gives exactly $2[\vec a\ \vec b\ \vec c]$. …

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4. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.The vector $\vec a$ lies in the plane of vectors $\vec b$ and $\vec c$, then (A) $[\vec a\ \vec b\ \vec c]$ (B) $[\vec a\ \vec b\ \vec c] = \vec b\ \vec c$ (C) $[\vec a\ \vec b\ \vec c] = 0$ (D) $[\vec a\ \vec b\ \vec c] = -1$

   ›Reveal solutionSolution

   Since $\vec a$ lying in the plane spanned by $\vec b,\vec c$ makes the three vectors coplanar, their scalar triple product must vanish.

   Concept and Intuition

   The scalar triple product $[\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times\vec c)$ measures the (signed) volume of the parallelepiped formed by the three vectors. If all three vectors lie in the same plane, that "parallelepiped" is flat — zero volume.

   Step-by-Step Solution

   1. If $\vec a$ lies in the plane of $\vec b$ and $\vec c$, then $\vec a$ can be written as a linear combination $\vec a=\lambda\vec b+\mu\vec c$ for some scalars $\lambda,\mu$.
   2. Compute $[\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times\vec c)=(\lambda\vec b+\mu\vec c)\cdot(\vec b\times\vec c)$.
   3. $\vec b\cdot(\vec b\times\vec c)=0$ (a vector is always perpendicular to a cross product involving itself), and similarly $\vec c\cdot(\vec b\times\vec c)=0$.
   4. So $[\vec a\ \vec b\ \vec c]=\lambda\cdot0+\mu\cdot0=0$.

   Common Mistakes …

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5. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.Let $\vec{a} = \hat{i} - \hat{j}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{c} = \hat{k} - \hat{i}$. If $\vec{d}$ is a unit vector such $\vec{a} \cdot \vec{b} = 0 = [\vec{b}\ \vec{c}\ \vec{d}]$ then $\vec{d}$ is (A) $\pm\dfrac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}$ (B) $\pm\dfrac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}}$ (C) $\pm\dfrac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$ (D) $\pm\dfrac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}}$

   ›Reveal solutionSolution

   Solving $\vec a\cdot\vec d=0$ together with the coplanarity condition $[\vec b\ \vec c\ \vec d]=0$ and unit-length normalization pins down $\vec d=\pm\dfrac{\hat i+\hat j-2\hat k}{\sqrt6}$.

   Concept and Intuition

   Two linear conditions (perpendicularity to $\vec a$, and lying in the plane of $\vec b,\vec c$ via a zero triple product) reduce the three unknown components of $\vec d$ to a single free parameter along a specific direction; the unit-vector condition then fixes the magnitude (up to sign).

   Step-by-Step Solution

   1. Let $\vec d = x\hat i+y\hat j+z\hat k$, with $x^2+y^2+z^2=1$.
   2. $\vec a=\hat i-\hat j$, so $\vec a\cdot\vec d = x-y=0 \Rightarrow x=y$.
   3. $\vec b=\hat j-\hat k=(0,1,-1)$, $\vec c=\hat k-\hat i=(-1,0,1)$. Compute $[\vec b\ \vec c\ \vec d] = \begin{vmatrix}0&1&-1\\-1&0&1\\x&y&z\end{vmatrix}$.
   4. Expand: $0(0\cdot z-1\cdot y) - 1(-1\cdot z-1\cdot x) + (-1)(-1\cdot y-0\cdot x) = (z+x)+y = x+y+z$.
   5. Set to zero: $x+y+z=0$. Using $x=y$: $2x+z=0 \Rightarrow z=-2x$. …

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6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If $\bar{a} = 2\bar{i}-3\bar{j}+4\bar{k}$, $\bar{b} = \bar{i}+2\bar{j}-\bar{k}$, $\bar{c} = 3\bar{i}-\bar{j}+2\bar{k}$ and $\bar{d} = \bar{i}+\bar{j}+\bar{k}$ are four vectors, then $(\bar{a}\times\bar{b})\times(\bar{c}\times\bar{d}) =$ (A) $17\bar{i} - 15\bar{j} + 9\bar{k}$ (B) $31\bar{i} - \bar{j} + 23\bar{k}$ (C) $17\bar{i} - \bar{j} + 23\bar{k}$ (D) $31\bar{i} - 15\bar{j} + 9\bar{k}$

   ›Reveal solutionSolution

   A vector triple-product identity turns $(\bar a\times\bar b)\times(\bar c\times\bar d)$ into a combination of $\bar a$ and $\bar b$ only, avoiding two separate cross products. The answer is $31\bar i-\bar j+23\bar k$.

   Concept and Intuition

   Computing $\bar a\times\bar b$ and $\bar c\times\bar d$ separately and then crossing the results works, but it's faster (and less error-prone) to treat $\bar c\times\bar d$ as a single vector $\bar C$ and apply the standard triple-product (BAC–CAB) identity:

   $(\bar A\times\bar B)\times\bar C=(\bar A\cdot\bar C)\,\bar B-(\bar B\cdot\bar C)\,\bar A.$

   This reduces the problem to one cross product ($\bar c\times\bar d$) and two dot products.

   Step-by-Step Solution

   1. Given $\bar a=(2,-3,4)$, $\bar b=(1,2,-1)$, $\bar c=(3,-1,2)$, $\bar d=(1,1,1)$.
   2. Compute $\bar c\times\bar d = (c_yd_z-c_zd_y,\ c_zd_x-c_xd_z,\ c_xd_y-c_yd_x)$ $=((-1)(1)-(2)(1),\ (2)(1)-(3)(1),\ (3)(1)-(-1)(1)) = (-3,-1,4)$.
   3. Apply the identity with $\bar A=\bar a,\bar B=\bar b,\bar C=\bar c\times\bar d$:

   $(\bar a\times\bar b)\times(\bar c\times\bar d)=(\bar a\cdot(\bar c\times\bar d))\,\bar b-(\bar b\cdot(\bar c\times\bar d))\,\bar a.$

   4. $\bar a\cdot(-3,-1,4)=2(-3)+(-3)(-1)+4(4)=-6+3+16=13$.
   5. $\bar b\cdot(-3,-1,4)=1(-3)+2(-1)+(-1)(4)=-3-2-4=-9$.
   6. So the result is $13\bar b-(-9)\bar a=9\bar a+13\bar b$. …

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7. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.If $\bar{a} = \bar{i}+\bar{j}+\bar{k}$, $\bar{c} = \bar{j}-\bar{k}$, $\bar{a}\times\bar{b} = \bar{c}$ and $\bar{a}.\bar{b} = 3$, then $\bar{b} =$ (A) $\dfrac{1}{3}(5\bar{i}+2\bar{j}+2\bar{k})$ (B) $\dfrac{1}{3}(2\bar{i}+5\bar{j}+2\bar{k})$ (C) $\dfrac{1}{3}(2\bar{i}+2\bar{j}+3\bar{k})$ (D) $\dfrac{1}{3}(2\bar{i}+5\bar{j}+5\bar{k})$

   ›Reveal solutionSolution

   This tests solving simultaneously for an unknown vector using both a cross-product constraint and a dot-product constraint. Answer: $\bar b=\frac13(5\bar i+2\bar j+2\bar k)$.

   Concept and Intuition

   A cross product equation gives us component-wise linear relations (since the cross product formula expands into a linear combination of the unknown's components), and a dot product equation gives one more linear relation. Together, three linear equations in the three unknown components ($x,y,z$ of $\bar b$) let us solve uniquely (generically) for $\bar b$.

   Step-by-Step Solution

   1. Let $\bar b = x\bar i+y\bar j+z\bar k$. Given $\bar a=\bar i+\bar j+\bar k$, compute $\bar a\times\bar b$ using the determinant formula:

   $\bar a\times\bar b = \begin{vmatrix}\bar i&\bar j&\bar k\\1&1&1\\x&y&z\end{vmatrix} = \bar i(1\cdot z-1\cdot y)-\bar j(1\cdot z-1\cdot x)+\bar k(1\cdot y-1\cdot x) = (z-y)\bar i+(x-z)\bar j+(y-x)\bar k.$

   2. Set this equal to $\bar c=\bar j-\bar k=(0,1,-1)$: matching components gives $z-y=0$, $x-z=1$, $y-x=-1$.
   3. From $z-y=0$: $z=y$. From $y-x=-1$: $x=y+1$. Check $x-z=1$: $(y+1)-y=1$ ✓ (automatically satisfied, so these two equations are consistent but not independent — we need the dot product for the third equation).
   4. Use $\bar a\cdot\bar b = x+y+z = 3$. Substitute $z=y$ and $x=y+1$: $(y+1)+y+y = 3 \Rightarrow 3y+1=3 \Rightarrow 3y=2 \Rightarrow y=\dfrac23$.
   5. Then $z=y=\dfrac23$ and $x=y+1=\dfrac53$. …

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8. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $\bar{a} = \bar{i} + \bar{j} + \bar{k}$, $\bar{a}.\bar{b} = 1$ and $\bar{a} \times \bar{b} = \bar{j} - \bar{k}$, then $\bar{b} =$ (A) $\bar{i} - \bar{j} + \bar{k}$ (B) $2\bar{j} - \bar{k}$ (C) $\bar{i}$ (D) $2\bar{i}$

   ›Reveal solutionSolution

   Solving the simultaneous conditions $\bar a\cdot\bar b=1$ and $\bar a\times\bar b=\bar j-\bar k$ for $\bar b=(x,y,z)$ gives $\bar b=\bar i$.

   Concept and Intuition

   A vector $\bar b$ is fully determined (given $\bar a\neq 0$) by its component along $\bar a$ (fixed by the dot product) together with its component perpendicular to $\bar a$ (fixed by the cross product, since $\bar a\times\bar b$ depends only on the perpendicular part of $\bar b$). Here we just solve the linear equations coordinate-wise.

   Step-by-Step Solution

   1. Let $\bar b=(x,y,z)$. With $\bar a=(1,1,1)$: $\bar a\times\bar b=\begin{vmatrix}\bar i&\bar j&\bar k\\1&1&1\\x&y&z\end{vmatrix}=(z-y)\bar i-(z-x)\bar j+(y-x)\bar k$.
   2. This must equal $\bar j-\bar k = (0,1,-1)$: so $z-y=0$, $x-z=1$, $y-x=-1$.
   3. From $z=y$ and $x=z+1$: substitute into $\bar a\cdot\bar b=x+y+z=1$: $(z+1)+z+z=1 \Rightarrow 3z+1=1 \Rightarrow z=0$. …

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9. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\bar a=\bar i-2\bar j-3\bar k, \bar b=-2\bar i+3\bar j+4\bar k, \bar c=5\bar i-4\bar j+3\bar k$ and $\bar d=3\bar i+\bar j+5\bar k$ are four vectors then $(\bar a\times \bar b)\times(\bar c\times \bar d)=$ (A) $18\bar i+6\bar j+30\bar k$ (B) $8\bar i-3\bar j+8\bar k$ (C) $19\bar i-5\bar j+21\bar k$ (D) $27\bar i-8\bar j+29\bar k$

   ›Reveal solutionSolution

   Compute the two cross products $\bar a\times\bar b$ and $\bar c\times\bar d$ first, then cross those results together to get $18\bar i+6\bar j+30\bar k$.

   Concept and Intuition

   A nested cross product like $(\bar a\times\bar b)\times(\bar c\times\bar d)$ is evaluated most reliably by direct component computation in two stages rather than trying to force a triple-product identity, since it avoids sign slips.

   Step-by-Step Solution

   1. $\bar a\times\bar b$ with $\bar a=(1,-2,-3)$, $\bar b=(-2,3,4)$: $\left(((-2)(4)-(-3)(3)),\,((-3)(-2)-(1)(4)),\,((1)(3)-(-2)(-2))\right)=(1,2,-1)$.
   2. $\bar c\times\bar d$ with $\bar c=(5,-4,3)$, $\bar d=(3,1,5)$: $\left(((-4)(5)-(3)(1)),\,((3)(3)-(5)(5)),\,((5)(1)-(-4)(3))\right)=(-23,-16,17)$. …

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10. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If $\bar{a} = \bar{i} - \bar{j} + \bar{k}$, $\bar{b} = \bar{i} + \bar{j} - 2\bar{k}$, $\bar{c} = 2\bar{i} - 3\bar{j} - \bar{k}$, $\bar{d} = 2\bar{i} + \bar{j} + \bar{k}$ are four vectors then $(\bar{a} \times \bar{c}) \times (\bar{b} \times \bar{d}) =$ (A) $2\bar{i} + 19\bar{j} - 11\bar{k}$ (B) $-8\bar{i} + 19\bar{j} - 29\bar{k}$ (C) $2\bar{i} + \bar{j} - 11\bar{k}$ (D) $-8\bar{i} + \bar{j} - 29\bar{k}$

    ›Reveal solutionSolution

    This tests direct computation of a double cross product; the answer is $-8\bar i+\bar j-29\bar k$.

    Concept and Intuition

    Rather than relying on a triple-product identity (which needs careful sign bookkeeping for a cross-of-crosses), it is often safer and just as fast to compute each intermediate cross product directly component-wise, then cross the two resulting vectors.

    Step-by-Step Solution

    1. $\bar a\times\bar c$ with $\bar a=(1,-1,1),\bar c=(2,-3,-1)$: $i$-comp $=(-1)(-1)-1(-3)=1+3=4$; $j$-comp $=-(1(-1)-1(2))=-(-1-2)=3$; $k$-comp $=1(-3)-(-1)(2)=-3+2=-1$. So $\bar a\times\bar c=(4,3,-1)$.
    2. $\bar b\times\bar d$ with $\bar b=(1,1,-2),\bar d=(2,1,1)$: $i$-comp $=1(1)-(-2)(1)=1+2=3$; $j$-comp $=-(1(1)-(-2)(2))=-(1+4)=-5$; $k$-comp $=1(1)-1(2)=-1$. So $\bar b\times\bar d=(3,-5,-1)$. …

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11. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.Let $\bar{a} = \bar{i}-2\bar{j}$, $\bar{b} = 2\bar{j}+3\bar{k}$, $\bar{c} = p\bar{i}+q\bar{j}$ and $\bar{d} = p\bar{j}-q\bar{k}$ be four vectors. If $(\bar{a}\times\bar{b}).\bar{c} = 3 = (\bar{a}\times\bar{b}).\bar{d}$, then $3p+q =$ (A) $0$ (B) $3$ (C) $-2$ (D) $6$

    ›Reveal solutionSolution

    Using $(\bar a\times\bar b)\cdot\bar c=3$ and $(\bar a\times\bar b)\cdot\bar d=3$ as two linear equations in $p,q$ gives $p=1,\,q=-3$, so $3p+q=0$.

    Concept and Intuition

    The scalar triple product $(\bar a\times\bar b)\cdot\bar v$ is linear in the components of $\bar v$. So once $\bar a\times\bar b$ is computed as a fixed vector, each condition given (dotting with $\bar c$ and with $\bar d$) becomes a simple linear equation in the unknowns $p,q$. Two independent linear equations in two unknowns can be solved directly.

    Step-by-Step Solution

    1. $\bar a = (1,-2,0)$, $\bar b = (0,2,3)$. Compute $\bar a\times\bar b = \begin{vmatrix}\bar i&\bar j&\bar k\\1&-2&0\\0&2&3\end{vmatrix} = \bar i[(-2)(3)-(0)(2)] - \bar j[(1)(3)-(0)(0)] + \bar k[(1)(2)-(-2)(0)] = (-6,-3,2)$.
    2. $\bar c = (p,q,0)$. Then $(\bar a\times\bar b)\cdot\bar c = -6p-3q+0 = 3 \Rightarrow -6p-3q=3 \Rightarrow 2p+q=-1$ ... (i)
    3. $\bar d = (0,p,-q)$. Then $(\bar a\times\bar b)\cdot\bar d = 0-3p-2q = 3 \Rightarrow -3p-2q=3 \Rightarrow 3p+2q=-3$ ... (ii) …

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12. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.If $\bar{a} = \bar{i} + \bar{j} + \bar{k}$, $\bar{b} = \bar{i} - \bar{j} + \bar{k}$ and $\bar{c} = \bar{i} + \bar{j} - \bar{k}$, then match the items of List-I with those of List-II. List-I: A. $[\bar{a}\ \bar{b}\ \bar{c}] =$ B. $|\bar{a} + \bar{b} + \bar{c}|^2 =$ C. Volume of the tetrahedron for which $\bar{a}, \bar{b}, \bar{c}$ are coterminus edges is D. $|(\bar{a} \times \bar{b}) \times (\bar{a} \times \bar{c})| =$ List-II: I. $4$ II. $11$ III. $\dfrac{2}{3}$ IV. $4\sqrt{3}$ V. $12$ The correct matching is (A) A-I, B-II, C-III, D-V (B) A-I, B-III, C-II, D-V (C) A-I, B-II, C-V, D-IV (D) A-I, B-II, C-III, D-IV

    ›Reveal solutionSolution

    Four independent vector-algebra computations (scalar triple product, sum-of-vectors magnitude, tetrahedron volume, a vector triple-product identity) with $\bar a=(1,1,1),\bar b=(1,-1,1),\bar c=(1,1,-1)$. All four match List-II items I, II, III, IV respectively — option (D).

    Concept and Intuition

    The scalar triple product $[\bar a\ \bar b\ \bar c]=\bar a\cdot(\bar b\times\bar c)$ measures the (signed) volume of the parallelepiped spanned by the three vectors; one-sixth of its absolute value is the volume of the tetrahedron with those vectors as coterminous edges. The double cross product identity

    $(\vec p\times\vec q)\times(\vec p\times\vec r) = [\vec p\ \vec q\ \vec r]\,\vec p$

    follows from the vector triple-product rule $\vec X\times(\vec Y\times\vec Z)=(\vec X\cdot\vec Z)\vec Y-(\vec X\cdot\vec Y)\vec Z$ applied with $\vec X=\vec p\times\vec q,\ \vec Y=\vec p,\ \vec Z=\vec r$ and noting $(\vec p\times\vec q)\cdot\vec p=0$.

    Step-by-Step Solution

    1. A. $[\bar a\ \bar b\ \bar c]=\begin{vmatrix}1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}=1(1-1)-1(-1-1)+1(1+1)=0+2+2=4$. So A → I (4).
    2. B. $\bar a+\bar b+\bar c=(1+1+1,\,1-1+1,\,1+1-1)=(3,1,1)$. $|\bar a+\bar b+\bar c|^2=3^2+1^2+1^2=9+1+1=11$. So B → II (11).
    3. C. Volume of the tetrahedron with coterminous edges $\bar a,\bar b,\bar c$ is $\dfrac16\left|[\bar a\ \bar b\ \bar c]\right|=\dfrac16(4)=\dfrac23$. So C → III (2/3).
    4. D. Using the identity $(\bar a\times\bar b)\times(\bar a\times\bar c)=[\bar a\ \bar b\ \bar c]\,\bar a$: …

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