Easy Practice — Set 1

Quick Recap — Vectors: Basics

• Magnitude: $|a\hat i+b\hat j+c\hat k|=\sqrt{a^2+b^2+c^2}$; unit vector $\hat a=\dfrac{\vec a}{|\vec a|}$.
• $\hat i\cdot\hat i=1$, $\hat i\cdot\hat j=0$; $\hat i\times\hat i=\vec0$, $\hat i\times\hat j=\hat k$, $\hat j\times\hat k=\hat i$, $\hat k\times\hat i=\hat j$.
• Addition is commutative; $\vec a\cdot\vec a=|\vec a|^2$; $\vec a\times\vec b=-(\vec b\times\vec a)$.
• The position vector of $(x,y,z)$ is $x\hat i+y\hat j+z\hat k$.

Beyond-NCERT JEE Formulae

Vector Triple Product. This product is a VECTOR, and with the bracket on the right it expands by the "BAC-CAB" rule:

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

With the bracket on the left the expansion is different:

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

[JEE Tip] The cross product is NOT associative: in general $\vec a\times(\vec b\times\vec c)\ne(\vec a\times\vec b)\times\vec c$. Middle-vector rule: the answer is always a linear combination of the two vectors INSIDE the bracket, so $\vec a\times(\vec b\times\vec c)$ lies in the plane of $\vec b$ and $\vec c$ (and is perpendicular to $\vec a$), while $(\vec a\times\vec b)\times\vec c$ lies in the plane of $\vec a$ and $\vec b$.

Triple Product of Cross Products. The box product of the three face-normals is the square of the box product of the edges:

$$[\vec a\times\vec b\ \ \vec b\times\vec c\ \ \vec c\times\vec a]=[\vec a\ \vec b\ \vec c]^2.$$

Being a perfect square it is never negative, and it vanishes exactly when $\vec a,\vec b,\vec c$ are coplanar.

Resolution of a Vector. Every $\vec a$ splits uniquely into a part parallel to $\vec b$ plus a part perpendicular to $\vec b$. With $\hat b=\dfrac{\vec b}{|\vec b|}$:

• component along $\vec b$ is $(\vec a\cdot\hat b)\,\hat b$, which also equals $\dfrac{\vec a\cdot\vec b}{|\vec b|^2}\,\vec b$;
• component perpendicular to $\vec b$ is $\vec a-(\vec a\cdot\hat b)\,\hat b$.

The two pieces add back to $\vec a$ and are mutually perpendicular, since their dot product is $0$.

Linear Independence. Three vectors $\vec a,\vec b,\vec c$ are linearly independent -- equivalently, non-coplanar -- if and only if

$$[\vec a\ \vec b\ \vec c]\ne0.$$

If $[\vec a\ \vec b\ \vec c]=0$ they are coplanar, hence linearly dependent. In 3-D space at most three vectors can be independent, so ANY four or more vectors are automatically linearly dependent.

Cyclic Scalar Triple Product. Cycling the letters $\vec a\to\vec b\to\vec c\to\vec a$ leaves the box product unchanged:

$$\vec a\cdot(\vec b\times\vec c)=\vec b\cdot(\vec c\times\vec a)=\vec c\cdot(\vec a\times\vec b).$$

Swapping any two of them reverses the sign, and the dot and cross may be interchanged, since $\vec a\cdot(\vec b\times\vec c)=(\vec a\times\vec b)\cdot\vec c$.

Solved Examples — Beyond-NCERT Formulae

Example 1: Vector triple product two ways. Let $\vec a=\hat i+2\hat j-\hat k$, $\vec b=2\hat i-\hat j+\hat k$ and $\vec c=\hat i+\hat j+2\hat k$; find $\vec a\times(\vec b\times\vec c)$.

Direct route. First $\vec b\times\vec c=-3\hat i-3\hat j+3\hat k$, and crossing with $\vec a$ gives

$$\vec a\times(\vec b\times\vec c)=3\hat i+0\hat j+3\hat k=3\hat i+3\hat k.$$

Formula route. Here $\vec a\cdot\vec c=1+2-2=1$ and $\vec a\cdot\vec b=2-2-1=-1$, so

$$\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\,\vec b-(\vec a\cdot\vec b)\,\vec c=\vec b+\vec c=3\hat i+3\hat k.$$

Both routes agree, and the answer is a combination of $\vec b$ and $\vec c$, confirming it lies in their plane.

Example 2: Non-associativity. For the same $\vec a,\vec b,\vec c$ compute $(\vec a\times\vec b)\times\vec c$. Using the left-bracket rule with $\vec a\cdot\vec c=1$ and $\vec b\cdot\vec c=2-1+2=3$,

$$(\vec a\times\vec b)\times\vec c=(\vec a\cdot\vec c)\,\vec b-(\vec b\cdot\vec c)\,\vec a=\vec b-3\vec a=-\hat i-7\hat j+4\hat k.$$

This differs from the $3\hat i+3\hat k$ of Example 1, so $\vec a\times(\vec b\times\vec c)\ne(\vec a\times\vec b)\times\vec c$: the cross product is not associative.

Example 3: Verify $[\vec a\times\vec b\ \ \vec b\times\vec c\ \ \vec c\times\vec a]=[\vec a\ \vec b\ \vec c]^2$. Take $\vec a=\hat i+\hat j$, $\vec b=\hat j+\hat k$ and $\vec c=\hat i+\hat k$. The edge box product is

$$[\vec a\ \vec b\ \vec c]=\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}=2.$$

The three cross products are $\vec a\times\vec b=\hat i-\hat j+\hat k$, $\vec b\times\vec c=\hat i+\hat j-\hat k$ and $\vec c\times\vec a=-\hat i+\hat j+\hat k$, so

$$[\vec a\times\vec b\ \ \vec b\times\vec c\ \ \vec c\times\vec a]=\begin{vmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{vmatrix}=4=2^2.$$

The identity checks out: the box product of the cross vectors is the square of the original.

Example 4: Resolve along and perpendicular. Resolve $\vec a=3\hat i+\hat j+\hat k$ relative to $\vec b=2\hat i+2\hat j+\hat k$. Since $|\vec b|=3$, we have $\hat b=\dfrac{1}{3}(2\hat i+2\hat j+\hat k)$ and $\vec a\cdot\vec b=6+2+1=9$, giving $\vec a\cdot\hat b=3$.

• Component along $\vec b$ is $(\vec a\cdot\hat b)\,\hat b=3\cdot\dfrac{1}{3}(2\hat i+2\hat j+\hat k)=2\hat i+2\hat j+\hat k$.
• Component perpendicular is $\vec a-(2\hat i+2\hat j+\hat k)=\hat i-\hat j$.

Check: $(\hat i-\hat j)\cdot\vec b=2-2+0=0$, and the two parts sum back to $\vec a$.

Example 5: Independence test. Read off the box product.

Set A: $\vec a=\hat i+2\hat j+3\hat k$, $\vec b=2\hat i+3\hat j+4\hat k$ and $\vec c=3\hat i+4\hat j+5\hat k$ give

$$[\vec a\ \vec b\ \vec c]=\begin{vmatrix}1&2&3\\2&3&4\\3&4&5\end{vmatrix}=0,$$

so Set A is linearly dependent -- indeed $\vec c=2\vec b-\vec a$.

Set B: $\vec a=\hat i+\hat j$, $\vec b=\hat j+\hat k$ and $\vec c=\hat i+\hat k$ give $[\vec a\ \vec b\ \vec c]=2\ne0$, so Set B is linearly independent, i.e. non-coplanar.

Example 6: Cyclic symmetry of the box product. For $\vec a=2\hat i+\hat j-\hat k$, $\vec b=\hat i-\hat j+2\hat k$ and $\vec c=3\hat i+2\hat j-\hat k$:

• $\vec b\times\vec c=-3\hat i+7\hat j+5\hat k$, so $\vec a\cdot(\vec b\times\vec c)=-6+7-5=-4$;
• $\vec c\times\vec a=-\hat i+\hat j-\hat k$, so $\vec b\cdot(\vec c\times\vec a)=-1-1-2=-4$;
• $\vec a\times\vec b=\hat i-5\hat j-3\hat k$, so $\vec c\cdot(\vec a\times\vec b)=3-10+3=-4$.

All three equal $-4$, confirming $\vec a\cdot(\vec b\times\vec c)=\vec b\cdot(\vec c\times\vec a)=\vec c\cdot(\vec a\times\vec b)$.

Example 7: Jacobi identity from the expansion. Prove $\vec a\times(\vec b\times\vec c)+\vec b\times(\vec c\times\vec a)+\vec c\times(\vec a\times\vec b)=\vec 0$. Expand every term by the middle-vector rule:

$$\begin{aligned}\vec a\times(\vec b\times\vec c)&=(\vec a\cdot\vec c)\,\vec b-(\vec a\cdot\vec b)\,\vec c,\\ \vec b\times(\vec c\times\vec a)&=(\vec b\cdot\vec a)\,\vec c-(\vec b\cdot\vec c)\,\vec a,\\ \vec c\times(\vec a\times\vec b)&=(\vec c\cdot\vec b)\,\vec a-(\vec c\cdot\vec a)\,\vec b.\end{aligned}$$

Adding, each $\vec a,\vec b,\vec c$ coefficient cancels in a pair, using $\vec a\cdot\vec b=\vec b\cdot\vec a$, so the sum is $\vec 0$. Numerically, with the Example 1 vectors the three terms are $3\hat i+3\hat k$, then $-4\hat i-7\hat j+\hat k$ and $\hat i+7\hat j-4\hat k$, which add to $\vec 0$.