Let a line L passing through the point (1, 1, 1) be perpendicular to both the vectors 2i + 2j + k and i + 2j + 2k. If P(a, b, c) is the foot of perpendicular from the origin on the line L, then the value of 34(a + b + c) is :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q15 – JEE-MAIN 2026

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Home Maths Vector AlgebraQuestion #504

Let a line L passing through the point

(1, 1, 1)

be perpendicular to both the vectors

2 \hat{i} + 2 \hat{j} + \hat{k}

and

\hat{i} + 2 \hat{j} + 2 \hat{k}

. If P(a, b, c) is the foot of perpendicular from the origin on the line L, then the value of

34 (a + b + c)

is :

Clue 1 of 3: Direction Vector of a Line

If a line is perpendicular to two vectors, its direction vector is parallel to the cross product of those two vectors.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine the Direction Vector of Line L✦ Active

The line L passes through $A (1, 1, 1)$ and is perpendicular to $\vec{v_{1}} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{v_{2}} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ . Therefore, its direction vector $\vec{d}$ is parallel to the cross product of $\vec{v_{1}}$ and $\vec{v_{2}}$ .

\vec{d} = \vec{v_{1}} \times \vec{v_{2}} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 1 & 2 & 2 \end{matrix} | = \hat{i} (4 - 2) - \hat{j} (4 - 1) + \hat{k} (4 - 2) = 2 \hat{i} - 3 \hat{j} + 2 \hat{k}

The equation of line L is $\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + t (2 \hat{i} - 3 \hat{j} + 2 \hat{k})$ . Any point P on the line L can be written as $P (1 + 2 t, 1 - 3 t, 1 + 2 t)$ .

Step 2: Find the Foot of the Perpendicular P(a, b, c)○ Expand

Let P(a, b, c) be the foot of the perpendicular from the origin $O (0, 0, 0)$ to line L. The vector $\vec{O P}$ must be perpendicular to the direction vector $\vec{d}$ of line L. So, $\vec{O P} \cdot \vec{d} = 0$ .

\vec{O P} = (1 + 2 t) \hat{i} + (1 - 3 t) \hat{j} + (1 + 2 t) \hat{k}

Applying the dot product condition:

(1 + 2 t) (2) + (1 - 3 t) (- 3) + (1 + 2 t) (2) = 0 2 + 4 t - 3 + 9 t + 2 + 4 t = 0 1 + 17 t = 0 t = - \frac{1}{17}

Substitute $t = - \frac{1}{17}$ into the coordinates of P:

a = 1 + 2 (- \frac{1}{17}) = 1 - \frac{2}{17} = \frac{15}{17} b = 1 - 3 (- \frac{1}{17}) = 1 + \frac{3}{17} = \frac{20}{17} c = 1 + 2 (- \frac{1}{17}) = 1 - \frac{2}{17} = \frac{15}{17}

💡 Teacher's Secret Hint

Ensure careful calculation of the cross product and the dot product to avoid sign errors.

Step 3: Calculate the Final Value○ Expand

Now, calculate $a + b + c$ :

a + b + c = \frac{15}{17} + \frac{20}{17} + \frac{15}{17} = \frac{15 + 20 + 15}{17} = \frac{50}{17}

Finally, calculate $34 (a + b + c)$ :

34 (a + b + c) = 34 \times \frac{50}{17} = 2 \times 50 = 100

💡 Teacher's Secret Hint

Double-check the arithmetic, especially when dealing with fractions.

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Maths Question 15 – JEE-MAIN 2026

Clue 1 of 3: Direction Vector of a Line

🧠 Full Solution Path