JEE Mathematics 2023 Question with Solution

Given: $\vec{a} = 6\hat{i} + 9\hat{j} + 12\hat{k}$, $\vec{b} = a\hat{i} + 11\hat{j} - 2\hat{k}$, and $\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$.

Find: $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k})$.

From the given cross product condition,

$$\vec{a} \times (\vec{c} - \vec{b}) = 0$$

so $\vec{a}$ is parallel to $\vec{c} - \vec{b}$. Therefore,

$$\vec{c} - \vec{b} = k\vec{a}$$

where $k$ is a scalar.

Hence,

$$\vec{c} = \vec{b} + k\vec{a} = (a + 6k)\hat{i} + (11 + 9k)\hat{j} + (-2 + 12k)\hat{k}$$

Using $\vec{a} \cdot \vec{c} = -12$,

$$6(a + 6k) + 9(11 + 9k) + 12(-2 + 12k) = -12$$

Using $\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5$,

$$(a + 6k) - 2(11 + 9k) + (-2 + 12k) = 5$$

Solving these, we get

$$\vec{c} = 23\hat{i} + 2\hat{j} - 14\hat{k}$$

Now,

$$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 - 14 = 11$$

Therefore, $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 11$.

Given: $\vec{c} = (a + 6k)\hat{i} + (11 + 9k)\hat{j} + (-2 + 12k)\hat{k}$.

Find: the required dot product.

From

$$(a + 6k) - 2(11 + 9k) + (-2 + 12k) = 5$$

we get

$$a + 6k - 22 - 18k - 2 + 12k = 5$$ $$a - 24 = 5$$ $$a = 29$$

Now use

$$6(a + 6k) + 9(11 + 9k) + 12(-2 + 12k) = -12$$

Substituting $a = 29$,

$$6(29 + 6k) + 9(11 + 9k) + 12(-2 + 12k) = -12$$ $$174 + 36k + 99 + 81k - 24 + 144k = -12$$ $$249 + 261k = -12$$ $$261k = -261$$ $$k = -1$$

Therefore,

$$\vec{c} = (29 - 6)\hat{i} + (11 - 9)\hat{j} + (-2 - 12)\hat{k}$$ $$\vec{c} = 23\hat{i} + 2\hat{j} - 14\hat{k}$$

Finally,

$$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 - 14 = 11$$

Therefore, the required numerical value is 11.