JEE Mathematics 2023 Question with Solution

Given:

• $\vec{a} = a\hat{i} + b\hat{j} + \hat{k}$
• $\vec{b} = 2\hat{i} - 3\hat{j} + 4\hat{k}$
• $3a + 2b = 7$
• $\dfrac{a}{b} = \dfrac{x}{2}$

Find: $x$

For two vectors to be perpendicular, their dot product must be zero.

Substitute the given vectors:

Simplify:

So,

Using the other given equation,

Solve the simultaneous equations. Multiply the first equation by $2$:

Multiply the second equation by $3$:

Add them:

Therefore,

Substitute into $3a + 2b = 7$:

Now use the ratio:

Hence,

Therefore, the value of $x$ is $1$.

Given: the two vectors are perpendicular, so their scalar product is zero.

Find: the value of $x$ in $\dfrac{a}{b} = \dfrac{x}{2}$.

The scalar product of corresponding components gives:

Now combine this with:

Eliminate $b$ by making the coefficients of $b$ equal and opposite:

Add:

Substitute back:

Now,

Compare with:

So,

Therefore, the final answer is $1$.