JAMIA 2026 — Mathematics PYQ

We can solve this problem systematically by assuming a general form for vector $\vec{B}$ and satisfying the two given conditions.

Let vector $\vec{B}$ be:

$$\vec{B}=x\hat{i}+y\hat{j}+z\hat{k}$$

We are given:

• $\vec{A}=\hat{i}-\hat{j}+\hat{k}$

• $\vec{C}=-\hat{i}-\hat{j}$

Step 1: Apply the Dot Product Condition ($\vec{A}\cdot \vec{B}=1$)

$$\vec{A}\cdot \vec{B}=(\hat{i}-\hat{j}+\hat{k})\cdot (x\hat{i}+y\hat{j}+z\hat{k})=1$$

$$(1)(x)+(-1)(y)+(1)(z)=1$$

$$x-y+z=1\text{— (Equation 1)}$$

Step 2: Apply the Cross Product Condition ($\vec{A}\times \vec{B}=\vec{C}$)

The cross product $\vec{A}\times \vec{B}$ is calculated using the determinant method:

$$\vec{A}\times \vec{B}=\left|\begin{array}{ccc}\hat{i}& amp;\hat{j}& amp;\hat{k}\\ 1& amp;-1& amp;1\\ x& amp;y& amp;z\end{array}\right|$$

Expanding along the first row:

$$\vec{A}\times \vec{B}=\hat{i}(-1\cdot z-1\cdot y)-\hat{j}(1\cdot z-1\cdot x)+\hat{k}(1\cdot y-(-1)\cdot x)$$

$$\vec{A}\times \vec{B}=(-z-y)\hat{i}+(x-z)\hat{j}+(x+y)\hat{k}$$

We know that $\vec{A}\times \vec{B}=\vec{C}=-1\hat{i}-1\hat{j}+0\hat{k}$. Comparing the components on both sides:

1. $\hat{i}$-component: $-y-z=-1⟹y+z=1\text{— (Equation 2)}$

2. $\hat{j}$-component: $x-z=-1\text{— (Equation 3)}$

3. $\hat{k}$-component: $x+y=0⟹x=-y\text{— (Equation 4)}$

Step 3: Solve the Linear Equations

From Equation 4, we have:

$$y=-x$$

Substitute $y=-x$ into Equation 2:

$$-x+z=1⟹z-x=1$$

Now compare this with Equation 3 ($x-z=-1$), which gives the same relationship. Let's use Equation 1 to find the unique values:

$$x-y+z=1$$

Substitute $y=-x$ and $z=x+1$ (from Equation 3) into Equation 1:

$$x-(-x)+(x+1)=1$$

$$x+x+x+1=1$$

$$3x=0$$

$$x=0$$

Now find $y$ and $z$:

• Since $y=-x⟹y=0$

• Since $z=x+1⟹z=0+1=1$

Step 4: Construct Vector $\vec{B}$

Substitute the values of $x$, $y$, and $z$ back into our assumption:

$$\vec{B}=0\hat{i}+0\hat{j}+1\hat{k}$$

$$\vec{B}=\hat{k}$$