JEE Mathematics 2026 Question with Solution

Given: $\vec{a}=2\hat{i}-5\hat{j}+5\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+3\hat{k}$, $2(\vec{a}\times\vec{c})+3(\vec{b}\times\vec{c})=\vec{0}$, and $(\vec{a}-\vec{b})\cdot\vec{c}=-97$.

Find: $|\vec{c}\times\hat{k}|^2$.

Using distributivity of cross product,

$$2(\vec{a}\times\vec{c})+3(\vec{b}\times\vec{c})=(2\vec{a}+3\vec{b})\times\vec{c}=\vec{0}$$

Therefore, $\vec{c}$ is parallel to $2\vec{a}+3\vec{b}$, so

$$\vec{c}=\lambda(2\vec{a}+3\vec{b})$$

Now,

$$2\vec{a}=4\hat{i}-10\hat{j}+10\hat{k}$$ $$3\vec{b}=3\hat{i}-3\hat{j}+9\hat{k}$$

Hence,

$$2\vec{a}+3\vec{b}=7\hat{i}-13\hat{j}+19\hat{k}$$

So,

$$\vec{c}=\lambda(7\hat{i}-13\hat{j}+19\hat{k})$$

Also,

$$\vec{a}-\vec{b}=(2-1)\hat{i}+(-5+1)\hat{j}+(5-3)\hat{k}=\hat{i}-4\hat{j}+2\hat{k}$$

Using the dot product condition,

$$(\vec{a}-\vec{b})\cdot\vec{c}=\lambda\big(7+52+38\big)=97\lambda$$

Given that $97\lambda=-97$, we get

$$\lambda=-1$$

Therefore,

$$\vec{c}=-7\hat{i}+13\hat{j}-19\hat{k}$$

Now compute the cross product with $\hat{k}$:

$$\vec{c}\times\hat{k}=13\hat{i}+7\hat{j}$$

Hence,

$$|\vec{c}\times\hat{k}|^2=13^2+7^2=169+49=218$$

Therefore, the correct option is B.

Given: $(2\vec{a}+3\vec{b})\times\vec{c}=\vec{0}$ and $(\vec{a}-\vec{b})\cdot\vec{c}=-97$.

Find: $|\vec{c}\times\hat{k}|^2$.

If $\vec{u}\times\vec{v}=\vec{0}$, then the vectors are parallel. So directly take

$$\vec{c}=\lambda(2\vec{a}+3\vec{b})$$

Now,

$$2\vec{a}+3\vec{b}=7\hat{i}-13\hat{j}+19\hat{k}$$

and

$$\vec{a}-\vec{b}=\hat{i}-4\hat{j}+2\hat{k}$$

Then,

$$(\vec{a}-\vec{b})\cdot(2\vec{a}+3\vec{b})=7+52+38=97$$

So,

$$(\vec{a}-\vec{b})\cdot\vec{c}=97\lambda=-97 \Rightarrow \lambda=-1$$

Thus,

$$\vec{c}=-7\hat{i}+13\hat{j}-19\hat{k}$$

Since for $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$,

$$|\vec{c}\times\hat{k}|^2=x^2+y^2$$

we get

$$|\vec{c}\times\hat{k}|^2=(-7)^2+13^2=49+169=218$$

Therefore, the correct option is B.