If for some \alpha and \beta in R, the intersection of the following three places x + 4y – 2z = 1 x + 7y – 5z = b x + 5y + \alpha z = 5 is a line in R3, then \alpha + \beta is equal to :

Matrices and Determinants PYQ — JEE Main 2020 · 9 Jan · Shift 1 · Q33 with Solution

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Correct answer: C

Interpret the question

We are given three planes in $\mathbb{R}^3$ : $x+4y-2z=1$ $x+7y-5z=\beta$ $x+5y+\alpha z=5$

For their intersection to be a line, the system must be consistent and have rank $=2$ . That means one equation must be a linear combination of the other two, both in coefficients and constants.

Write the coefficient matrix

The coefficient rows are: $R_1=(1,4,-2),\quad R_2=(1,7,-5),\quad R_3=(1,5,\alpha)$

For the three planes to intersect in a line, these three rows must be linearly dependent:

1&4&-2\\ 1&7&-5\\ 1&5&\alpha \end{pmatrix}=0$$ Compute the determinant: $$\begin{vmatrix} 1&4&-2\\ 1&7&-5\\ 1&5&\alpha \end{vmatrix} =1\begin{vmatrix}7&-5\\5&\alpha\end{vmatrix}-4\begin{vmatrix}1&-5\\1&\alpha\end{vmatrix}+(-2)\begin{vmatrix}1&7\\1&5\end{vmatrix}$$ $$=(7\alpha+25)-4(\alpha+5)+(-2)(5-7)$$ $$=7\alpha+25-4\alpha-20+4$$ $$=3\alpha+9$$ Set this equal to zero: $$3\alpha+9=0$$ $$\alpha=-3$$ --- 3. **Find the relation among the planes** Now with $\alpha=-3$, the third plane is: $$x+5y-3z=5$$ Let us see if its coefficient row is a combination of the first two: Assume $$R_3=mR_1+nR_2$$ Then $$(1,5,-3)=m(1,4,-2)+n(1,7,-5)$$ So, $$m+n=1$$ $$4m+7n=5$$ $$-2m-5n=-3$$ From $m+n=1$, take $m=1-n$. Substitute into $4m+7n=5$: $$4(1-n)+7n=5$$ $$4+3n=5$$ $$n=\frac13,\, m=\frac23$$ Check third equation: $$-2\left(\frac23\right)-5\left(\frac13\right)=-\frac43-\frac53=-3$$ Correct. Thus, $$R_3=\frac23R_1+\frac13R_2$$ For consistency and intersection as a line, the constants must satisfy the same relation: $$5=\frac23(1)+\frac13(\beta)$$ Multiply by $3$: $$15=2+\beta$$ $$\beta=13$$ --- 4. **Compute $\alpha+\beta$** $$\alpha+\beta=-3+13=10$$ --- 5. **Check with options** Option C is: $$10$$ So the correct answer is **C**.

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View written solutionFree

Correct answer: C

Interpret the question

We are given three planes in $\mathbb{R}^3$ : $x+4y-2z=1$ $x+7y-5z=\beta$ $x+5y+\alpha z=5$

For their intersection to be a line, the system must be consistent and have rank $=2$ . That means one equation must be a linear combination of the other two, both in coefficients and constants.

Write the coefficient matrix

The coefficient rows are: $R_1=(1,4,-2),\quad R_2=(1,7,-5),\quad R_3=(1,5,\alpha)$

For the three planes to intersect in a line, these three rows must be linearly dependent:

1&4&-2\\ 1&7&-5\\ 1&5&\alpha \end{pmatrix}=0$$ Compute the determinant: $$\begin{vmatrix} 1&4&-2\\ 1&7&-5\\ 1&5&\alpha \end{vmatrix} =1\begin{vmatrix}7&-5\\5&\alpha\end{vmatrix}-4\begin{vmatrix}1&-5\\1&\alpha\end{vmatrix}+(-2)\begin{vmatrix}1&7\\1&5\end{vmatrix}$$ $$=(7\alpha+25)-4(\alpha+5)+(-2)(5-7)$$ $$=7\alpha+25-4\alpha-20+4$$ $$=3\alpha+9$$ Set this equal to zero: $$3\alpha+9=0$$ $$\alpha=-3$$ --- 3. **Find the relation among the planes** Now with $\alpha=-3$, the third plane is: $$x+5y-3z=5$$ Let us see if its coefficient row is a combination of the first two: Assume $$R_3=mR_1+nR_2$$ Then $$(1,5,-3)=m(1,4,-2)+n(1,7,-5)$$ So, $$m+n=1$$ $$4m+7n=5$$ $$-2m-5n=-3$$ From $m+n=1$, take $m=1-n$. Substitute into $4m+7n=5$: $$4(1-n)+7n=5$$ $$4+3n=5$$ $$n=\frac13,\, m=\frac23$$ Check third equation: $$-2\left(\frac23\right)-5\left(\frac13\right)=-\frac43-\frac53=-3$$ Correct. Thus, $$R_3=\frac23R_1+\frac13R_2$$ For consistency and intersection as a line, the constants must satisfy the same relation: $$5=\frac23(1)+\frac13(\beta)$$ Multiply by $3$: $$15=2+\beta$$ $$\beta=13$$ --- 4. **Compute $\alpha+\beta$** $$\alpha+\beta=-3+13=10$$ --- 5. **Check with options** Option C is: $$10$$ So the correct answer is **C**.

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